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Tangent lines practice problems

The equation of the line tangent to a curve at a point. 70 problems with worked solutions; in 70 of them every equation is proved by a computer algebra system.

20 worked problems

Try each one before opening the solution. Every line is labeled with what checked it. How we verify

Problem 2.725

medium Video
Find an equation of the tangent line to \( \displaystyle y = 2 x + \sin{\left(x \right)} \) at \( \displaystyle x = \pi \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. 2 x + \sin{\left(x \right)} \right|_{\substack{ x=\pi }} = 2 \pi \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} \left(2 x + \sin{\left(x \right)}\right) = \cos{\left(x \right)} + 2 \]
    Differentiate.✓ Proved
  4. \[ \left. \cos{\left(x \right)} + 2 \right|_{\substack{ x=\pi }} = 1 \]
    The slope at the point.✓ Proved
  5. \[ x + \pi \]
    Point-slope form, then simplify.Reviewed
Answer \( y = x + \pi \)
The graph of y = 2x + sin(x) near x = pi. A secant line through the point (pi, 2pi) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at 1, giving y = x + pi.

✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.726

medium Video
Find an equation of the tangent line to \( \displaystyle y = x e^{2 x} \) at \( \displaystyle x = 2 \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. x e^{2 x} \right|_{\substack{ x=2 }} = 2 e^{4} \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} x e^{2 x} = \left(2 x + 1\right) e^{2 x} \]
    Differentiate.✓ Proved
  4. \[ \left. \left(2 x + 1\right) e^{2 x} \right|_{\substack{ x=2 }} = 5 e^{4} \]
    The slope at the point.✓ Proved
  5. \[ \left(5 x - 10\right) e^{4} + 2 e^{4} = 5 x e^{4} - 8 e^{4} \]
    Point-slope form, then simplify.✓ Proved
Answer \( y = 5 x e^{4} - 8 e^{4} \)
The graph of y = x·e^(2x) near x = 2. A secant line through the point (2, 2e^(4)) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at 5e^(4), giving y = 5x·e^(4) - 8e^(4).

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.728

medium Video
Find an equation of the tangent line to \( \displaystyle y = - x - 2 \sin{\left(x \right)} + 2 \) at \( \displaystyle x = \pi \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. - x - 2 \sin{\left(x \right)} + 2 \right|_{\substack{ x=\pi }} = 2 - \pi \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} \left(- x - 2 \sin{\left(x \right)} + 2\right) = - 2 \cos{\left(x \right)} - 1 \]
    Differentiate.✓ Proved
  4. \[ \left. - 2 \cos{\left(x \right)} - 1 \right|_{\substack{ x=\pi }} = 1 \]
    The slope at the point.✓ Proved
  5. \[ x - 2 \pi + 2 \]
    Point-slope form, then simplify.Reviewed
Answer \( y = x - 2 \pi + 2 \)
The graph of y = -x - 2sin(x) + 2 near x = pi. A secant line through the point (pi, 2 - pi) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at 1, giving y = x - 2pi + 2.

✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.731

medium Video
Find an equation of the tangent line to \( \displaystyle y = x e^{- x} \) at \( \displaystyle x = -1 \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. x e^{- x} \right|_{\substack{ x=-1 }} = - e \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} x e^{- x} = \left(1 - x\right) e^{- x} \]
    Differentiate.✓ Proved
  4. \[ \left. \left(1 - x\right) e^{- x} \right|_{\substack{ x=-1 }} = 2 e \]
    The slope at the point.✓ Proved
  5. \[ 2 e \left(x + 1\right) - e = 2 e x + e \]
    Point-slope form, then simplify.✓ Proved
Answer \( y = 2 e x + e \)
The graph of y = x·e^(-x) near x = -1. A secant line through the point (-1, -E) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at 2E, giving y = 2E·x + E.

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.732

medium Video
Find an equation of the tangent line to \( \displaystyle y = x e^{2 x} \) at \( \displaystyle x = -1 \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. x e^{2 x} \right|_{\substack{ x=-1 }} = - \frac{1}{e^{2}} \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} x e^{2 x} = \left(2 x + 1\right) e^{2 x} \]
    Differentiate.✓ Proved
  4. \[ \left. \left(2 x + 1\right) e^{2 x} \right|_{\substack{ x=-1 }} = - \frac{1}{e^{2}} \]
    The slope at the point.✓ Proved
  5. \[ \frac{- x - 1}{e^{2}} - e^{-2} = - \frac{x}{e^{2}} - \frac{2}{e^{2}} \]
    Point-slope form, then simplify.✓ Proved
Answer \( y = - \frac{x}{e^{2}} - \frac{2}{e^{2}} \)
The graph of y = x·e^(2x) near x = -1. A secant line through the point (-1, -e^(-2)) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at -e^(-2), giving y = -x·e^(-2) - 2e^(-2).

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.733

medium Video
Find an equation of the tangent line to \( \displaystyle y = - 2 x^{2} + x + 3 - \frac{3}{x} \) at \( \displaystyle x = -1 \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. - 2 x^{2} + x + 3 - \frac{3}{x} \right|_{\substack{ x=-1 }} = 3 \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} \left(- 2 x^{2} + x + 3 - \frac{3}{x}\right) = - 4 x + 1 + \frac{3}{x^{2}} \]
    Differentiate.✓ Proved
  4. \[ \left. - 4 x + 1 + \frac{3}{x^{2}} \right|_{\substack{ x=-1 }} = 8 \]
    The slope at the point.✓ Proved
  5. \[ 8 x + 11 \]
    Point-slope form, then simplify.Reviewed
Answer \( y = 8 x + 11 \)
The graph of y = -2x² + x + 3 - 3/x near x = -1. A secant line through the point (-1, 3) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at 8, giving y = 8x + 11.

✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.734

medium Video
Find an equation of the tangent line to \( \displaystyle y = x e^{- 2 x} \) at \( \displaystyle x = -1 \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. x e^{- 2 x} \right|_{\substack{ x=-1 }} = - e^{2} \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} x e^{- 2 x} = \left(1 - 2 x\right) e^{- 2 x} \]
    Differentiate.✓ Proved
  4. \[ \left. \left(1 - 2 x\right) e^{- 2 x} \right|_{\substack{ x=-1 }} = 3 e^{2} \]
    The slope at the point.✓ Proved
  5. \[ \left(3 x + 3\right) e^{2} - e^{2} = 3 x e^{2} + 2 e^{2} \]
    Point-slope form, then simplify.✓ Proved
Answer \( y = 3 x e^{2} + 2 e^{2} \)
The graph of y = x·e^(-2x) near x = -1. A secant line through the point (-1, -e^(2)) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at 3e^(2), giving y = 3x·e^(2) + 2e^(2).

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.738

medium Video
Find an equation of the tangent line to \( \displaystyle y = - x + 3 \sin{\left(x \right)} \) at \( \displaystyle x = \pi \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. - x + 3 \sin{\left(x \right)} \right|_{\substack{ x=\pi }} = - \pi \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} \left(- x + 3 \sin{\left(x \right)}\right) = 3 \cos{\left(x \right)} - 1 \]
    Differentiate.✓ Proved
  4. \[ \left. 3 \cos{\left(x \right)} - 1 \right|_{\substack{ x=\pi }} = -4 \]
    The slope at the point.✓ Proved
  5. \[ - 4 x + 3 \pi \]
    Point-slope form, then simplify.Reviewed
Answer \( y = - 4 x + 3 \pi \)
The graph of y = -x + 3sin(x) near x = pi. A secant line through the point (pi, -pi) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at -4, giving y = -4x + 3pi.

✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.772

medium Video
Find an equation of the tangent line to \( \displaystyle y = - x^{2} - 2 x - 4 - \frac{3}{x} \) at \( \displaystyle x = 2 \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. - x^{2} - 2 x - 4 - \frac{3}{x} \right|_{\substack{ x=2 }} = - \frac{27}{2} \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} \left(- x^{2} - 2 x - 4 - \frac{3}{x}\right) = - 2 x - 2 + \frac{3}{x^{2}} \]
    Differentiate.✓ Proved
  4. \[ \left. - 2 x - 2 + \frac{3}{x^{2}} \right|_{\substack{ x=2 }} = - \frac{21}{4} \]
    The slope at the point.✓ Proved
  5. \[ - \frac{21 x}{4} - 3 \]
    Point-slope form, then simplify.Reviewed
Answer \( y = - \frac{21 x}{4} - 3 \)
The graph of y = -x² - 2x - 4 - 3/x near x = 2. A secant line through the point (2, -27/2) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at -21/4, giving y = -21x/4 - 3.

✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.773

medium Video
Find an equation of the tangent line to \( \displaystyle y = 3 x^{2} - 3 x + 2 + \frac{2}{x} \) at \( \displaystyle x = -1 \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. 3 x^{2} - 3 x + 2 + \frac{2}{x} \right|_{\substack{ x=-1 }} = 6 \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} \left(3 x^{2} - 3 x + 2 + \frac{2}{x}\right) = 6 x - 3 - \frac{2}{x^{2}} \]
    Differentiate.✓ Proved
  4. \[ \left. 6 x - 3 - \frac{2}{x^{2}} \right|_{\substack{ x=-1 }} = -11 \]
    The slope at the point.✓ Proved
  5. \[ - 11 x - 5 \]
    Point-slope form, then simplify.Reviewed
Answer \( y = - 11 x - 5 \)
The graph of y = 3x² - 3x + 2 + 2/x near x = -1. A secant line through the point (-1, 6) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at -11, giving y = -11x - 5.

✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.776

medium Video
Find an equation of the tangent line to \( \displaystyle y = x e^{x} \) at \( \displaystyle x = 4 \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. x e^{x} \right|_{\substack{ x=4 }} = 4 e^{4} \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} x e^{x} = \left(x + 1\right) e^{x} \]
    Differentiate.✓ Proved
  4. \[ \left. \left(x + 1\right) e^{x} \right|_{\substack{ x=4 }} = 5 e^{4} \]
    The slope at the point.✓ Proved
  5. \[ \left(5 x - 20\right) e^{4} + 4 e^{4} = 5 x e^{4} - 16 e^{4} \]
    Point-slope form, then simplify.✓ Proved
Answer \( y = 5 x e^{4} - 16 e^{4} \)
The graph of y = x·e^(x) near x = 4. A secant line through the point (4, 4e^(4)) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at 5e^(4), giving y = 5x·e^(4) - 16e^(4).

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.778

medium Video
Find an equation of the tangent line to \( \displaystyle y = x e^{- 2 x} \) at \( \displaystyle x = 2 \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. x e^{- 2 x} \right|_{\substack{ x=2 }} = \frac{2}{e^{4}} \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} x e^{- 2 x} = \left(1 - 2 x\right) e^{- 2 x} \]
    Differentiate.✓ Proved
  4. \[ \left. \left(1 - 2 x\right) e^{- 2 x} \right|_{\substack{ x=2 }} = - \frac{3}{e^{4}} \]
    The slope at the point.✓ Proved
  5. \[ \frac{6 - 3 x}{e^{4}} + \frac{2}{e^{4}} = - \frac{3 x}{e^{4}} + \frac{8}{e^{4}} \]
    Point-slope form, then simplify.✓ Proved
Answer \( y = - \frac{3 x}{e^{4}} + \frac{8}{e^{4}} \)
The graph of y = x·e^(-2x) near x = 2. A secant line through the point (2, 2e^(-4)) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at -3e^(-4), giving y = -3x·e^(-4) + 8e^(-4).

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.911

medium Video
Find an equation of the tangent line to \( \displaystyle y = \sqrt{x} - x + 3 \) at \( \displaystyle x = 1 \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. \sqrt{x} - x + 3 \right|_{\substack{ x=1 }} = 3 \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} \left(\sqrt{x} - x + 3\right) = -1 + \frac{1}{2 \sqrt{x}} \]
    Differentiate.✓ Proved
  4. \[ \left. -1 + 1 \frac{1}{2 \sqrt{x}} \right|_{\substack{ x=1 }} = - \frac{1}{2} \]
    The slope at the point.✓ Proved
  5. \[ - \frac{x}{2} + \frac{7}{2} = \frac{7}{2} - \frac{x}{2} \]
    Point-slope form, then simplify.Reviewed
Answer \( y = \frac{7}{2} - \frac{x}{2} \)
The graph of y = sqrt(x) - x + 3 near x = 1. A secant line through the point (1, 3) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at -1/2, giving y = 7/2 - x/2.

✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.912

medium Video
Find an equation of the tangent line to \( \displaystyle y = - x^{2} + 4 x + 3 + \frac{3}{x} \) at \( \displaystyle x = 1 \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. - x^{2} + 4 x + 3 + \frac{3}{x} \right|_{\substack{ x=1 }} = 9 \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} \left(- x^{2} + 4 x + 3 + \frac{3}{x}\right) = - 2 x + 4 - \frac{3}{x^{2}} \]
    Differentiate.✓ Proved
  4. \[ \left. - 2 x + 4 - \frac{3}{x^{2}} \right|_{\substack{ x=1 }} = -1 \]
    The slope at the point.✓ Proved
  5. \[ 10 - x \]
    Point-slope form, then simplify.Reviewed
Answer \( y = 10 - x \)
The graph of y = -x² + 4x + 3 + 3/x near x = 1. A secant line through the point (1, 9) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at -1, giving y = 10 - x.

✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.917

medium Video
Find an equation of the tangent line to \( \displaystyle y = x + 3 \sin{\left(x \right)} + 2 \) at \( \displaystyle x = 0 \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. x + 3 \sin{\left(x \right)} + 2 \right|_{\substack{ x=0 }} = 2 \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} \left(x + 3 \sin{\left(x \right)} + 2\right) = 3 \cos{\left(x \right)} + 1 \]
    Differentiate.✓ Proved
  4. \[ \left. 3 \cos{\left(x \right)} + 1 \right|_{\substack{ x=0 }} = 4 \]
    The slope at the point.✓ Proved
  5. \[ 4 x + 2 \]
    Point-slope form, then simplify.Reviewed
Answer \( y = 4 x + 2 \)
The graph of y = x + 3sin(x) + 2 near x = 0. A secant line through the point (0, 2) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at 4, giving y = 4x + 2.

✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.1037

medium Video
Find an equation of the tangent line to \( \displaystyle y = 2 \sqrt{x} + x - 1 \) at \( \displaystyle x = 4 \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. 2 \sqrt{x} + x - 1 \right|_{\substack{ x=4 }} = 7 \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} \left(2 \sqrt{x} + x - 1\right) = 1 + \frac{1}{\sqrt{x}} \]
    Differentiate.✓ Proved
  4. \[ \left. 1 + 1 \frac{1}{\sqrt{x}} \right|_{\substack{ x=4 }} = \frac{3}{2} \]
    The slope at the point.✓ Proved
  5. \[ \frac{3 x}{2} + 1 \]
    Point-slope form, then simplify.Reviewed
Answer \( y = \frac{3 x}{2} + 1 \)
The graph of y = 2sqrt(x) + x - 1 near x = 4. A secant line through the point (4, 7) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at 3/2, giving y = 3x/2 + 1.

✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.1041

medium Video
Find an equation of the tangent line to \( \displaystyle y = \sqrt{x} + x - 3 \) at \( \displaystyle x = 4 \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. \sqrt{x} + x - 3 \right|_{\substack{ x=4 }} = 3 \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} \left(\sqrt{x} + x - 3\right) = 1 + \frac{1}{2 \sqrt{x}} \]
    Differentiate.✓ Proved
  4. \[ \left. 1 + 1 \frac{1}{2 \sqrt{x}} \right|_{\substack{ x=4 }} = \frac{5}{4} \]
    The slope at the point.✓ Proved
  5. \[ \frac{5 x}{4} - 2 \]
    Point-slope form, then simplify.Reviewed
Answer \( y = \frac{5 x}{4} - 2 \)
The graph of y = sqrt(x) + x - 3 near x = 4. A secant line through the point (4, 3) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at 5/4, giving y = 5x/4 - 2.

✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.1042

medium Video
Find an equation of the tangent line to \( \displaystyle y = - 2 x - \sin{\left(x \right)} - 2 \) at \( \displaystyle x = \pi \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. - 2 x - \sin{\left(x \right)} - 2 \right|_{\substack{ x=\pi }} = - 2 \pi - 2 \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} \left(- 2 x - \sin{\left(x \right)} - 2\right) = - \cos{\left(x \right)} - 2 \]
    Differentiate.✓ Proved
  4. \[ \left. - \cos{\left(x \right)} - 2 \right|_{\substack{ x=\pi }} = -1 \]
    The slope at the point.✓ Proved
  5. \[ - x - \pi - 2 \]
    Point-slope form, then simplify.Reviewed
Answer \( y = - x - \pi - 2 \)
The graph of y = -2x - sin(x) - 2 near x = pi. A secant line through the point (pi, -2pi - 2) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at -1, giving y = -x - pi - 2.

✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.1043

medium Video
Find an equation of the tangent line to \( \displaystyle y = - 3 x^{2} - x + 2 + \frac{3}{x} \) at \( \displaystyle x = 1 \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. - 3 x^{2} - x + 2 + \frac{3}{x} \right|_{\substack{ x=1 }} = 1 \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} \left(- 3 x^{2} - x + 2 + \frac{3}{x}\right) = - 6 x - 1 - \frac{3}{x^{2}} \]
    Differentiate.✓ Proved
  4. \[ \left. - 6 x - 1 - \frac{3}{x^{2}} \right|_{\substack{ x=1 }} = -10 \]
    The slope at the point.✓ Proved
  5. \[ 11 - 10 x \]
    Point-slope form, then simplify.Reviewed
Answer \( y = 11 - 10 x \)
The graph of y = -3x² - x + 2 + 3/x near x = 1. A secant line through the point (1, 1) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at -10, giving y = 11 - 10x.

✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 2.1152

medium Video
Find an equation of the tangent line to \( \displaystyle y = x e^{- x} \) at \( \displaystyle x = 2 \).
  1. The tangent line passes through the point (a, f(a)) and has slope f'(a).
    Reviewed
  2. \[ \left. x e^{- x} \right|_{\substack{ x=2 }} = \frac{2}{e^{2}} \]
    The point of tangency.✓ Proved
  3. \[ \frac{d}{d x} x e^{- x} = \left(1 - x\right) e^{- x} \]
    Differentiate.✓ Proved
  4. \[ \left. \left(1 - x\right) e^{- x} \right|_{\substack{ x=2 }} = - \frac{1}{e^{2}} \]
    The slope at the point.✓ Proved
  5. \[ \frac{2 - x}{e^{2}} + \frac{2}{e^{2}} = - \frac{x}{e^{2}} + \frac{4}{e^{2}} \]
    Point-slope form, then simplify.✓ Proved
Answer \( y = - \frac{x}{e^{2}} + \frac{4}{e^{2}} \)
The graph of y = x·e^(-x) near x = 2. A secant line through the point (2, 2e^(-2)) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at -e^(-2), giving y = -x·e^(-2) + 4e^(-2).

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

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