Question

finding the slope of a graph 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\t\tpoint

1. ( f(x) = \frac{8}{x^2} )\t\t( (2, 2) )
2. ( f(t) = 2 - \frac{4}{t} )\t\t( (4, 1) )
3. ( f(x) = -\frac{1}{2} + \frac{7}{5}x^3 )\t\t( left(0, -\frac{1}{2}

ight) )

1. ( y = 2x^4 - 3 )\t\t( (1, -1) )
2. ( y = (4x + 1)^2 )\t\t( (0, 1) )

Explanation:

Problem 31:

Step1: Rewrite the function

Rewrite \( f(x)=\frac{8}{x^2} \) as \( f(x) = 8x^{-2} \).

Step2: Find the derivative

Using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \), the derivative \( f^\prime(x)=8\times(-2)x^{-2-1}=-16x^{-3}=-\frac{16}{x^3} \).

Step3: Evaluate at \( x = 2 \)

Substitute \( x = 2 \) into \( f^\prime(x) \): \( f^\prime(2)=-\frac{16}{2^3}=-\frac{16}{8}=-2 \).

Step1: Rewrite the function

Rewrite \( f(t)=2-\frac{4}{t} \) as \( f(t)=2 - 4t^{-1} \).

Step2: Find the derivative

Using the power rule, \( f^\prime(t)=0-4\times(-1)t^{-1 - 1}=4t^{-2}=\frac{4}{t^2} \).

Step3: Evaluate at \( t = 4 \)

Substitute \( t = 4 \) into \( f^\prime(t) \): \( f^\prime(4)=\frac{4}{4^2}=\frac{4}{16}=\frac{1}{4} \).

Step1: Find the derivative

Using the power rule, for \( f(x)=-\frac{1}{2}+\frac{7}{5}x^3 \), \( f^\prime(x)=0+\frac{7}{5}\times3x^{3 - 1}=\frac{21}{5}x^2 \).

Step2: Evaluate at \( x = 0 \)

Substitute \( x = 0 \) into \( f^\prime(x) \): \( f^\prime(0)=\frac{21}{5}\times0^2 = 0 \).

Answer:

-2

Problem 32: