Question

In Exercises 31–38, find the slope of the graph of the function at the given point. Use the derivative feature of a graphing utility to confirm your results. Function Point 31. $f(x) = frac{8}{x^2}$ (2, 2) 32. $f(t) = 3 - frac{3}{5t}$ $(frac{3}{5}, 2)$ 33. $f(x) = -frac{1}{2} + frac{7}{5}x^3$ $(0, -frac{1}{2})$ 34. $y = 3x^3 - 10$ (2, 14)

          In Exercises 31–38, find the slope of the graph of the function at the given point. Use the derivative feature of a graphing utility to confirm your results.
Function
Point
31. $f(x) = frac{8}{x^2}$ (2, 2)
32. $f(t) = 3 - frac{3}{5t}$ $(frac{3}{5}, 2)$
33. $f(x) = -frac{1}{2} + frac{7}{5}x^3$ $(0, -frac{1}{2})$
34. $y = 3x^3 - 10$ (2, 14)

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$In Exercises 31–38, find the slope of the graph of the function at the given point. Use the derivative feature of a graphing utility to confirm your results. Function Point 31. f(x) = frac8x^2 (2, 2) 32. f(t) = 3 - frac35t (frac35, 2) 33. f(x) = -frac12 + frac75x^3 (0, -frac12) 34. y = 3x^3 - 10 (2, 14)$

Added by Michael C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

06/06/2022

Step 1

The derivative of \( f(x) = x^2 \) is \( f'(x) = 2x \). Plugging in x = 2, we get \( f'(2) = 2(2) = 4 \). Therefore, the slope at the point (2,2) is 4. Show more…

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In Exercises 31-38, find the slope of the graph of the function at the given point: Use the derivative feature of a graphing utility to confirm your results. Function Point 31. f(x) = 8/x^2 (2, 2) 32. f(t) = 3 - 3/(5t) (3/5, 2) 33. f(x) = -1/2 + 7/5x^3 (0, -1/2) 34. y = 3x^3 - 10 (2, 14)

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