Question

introduction to derivatives: problem 1 (1 point) let ( f(x) = x^3 - 4x ). calculate the difference quotient ( \frac{f(3+h) - f(3)}{h} ) for ( h = .1 ), ( h = .01 ), ( h = -.01 ), ( h = -.1 ). if someone now told you that the derivative (slope of the tangent line to the graph) of ( f(x) ) at ( x = 3 ) was an integer, what would you expect it to be?

Explanation:

For \( h = 0.1 \):

Step1: Calculate \( f(3 + h) \) and \( f(3) \)

First, find \( f(3) \): \( f(3)=3^{3}-4\times3 = 27 - 12 = 15 \).
Then, \( f(3 + 0.1)=f(3.1)=(3.1)^{3}-4\times3.1 = 29.791 - 12.4 = 17.391 \).

Step2: Compute the difference quotient

The difference quotient is \( \frac{f(3 + h)-f(3)}{h} \). Substitute \( h = 0.1 \), \( f(3 + 0.1)=17.391 \), and \( f(3)=15 \):
\( \frac{17.391 - 15}{0.1}=\frac{2.391}{0.1}=23.91 \).

For \( h = 0.01 \):

Step1: Calculate \( f(3 + h) \) and \( f(3) \)

\( f(3)=15 \) (from before).
\( f(3 + 0.01)=f(3.01)=(3.01)^{3}-4\times3.01 = 27.270901 - 12.04 = 15.230901 \).

Step2: Compute the difference quotient

Substitute into \( \frac{f(3 + h)-f(3)}{h} \):
\( \frac{15.230901 - 15}{0.01}=\frac{0.230901}{0.01}=23.0901 \).

For \( h = -0.01 \):

Step1: Calculate \( f(3 + h) \) and \( f(3) \)

\( f(3)=15 \).
\( f(3 - 0.01)=f(2.99)=(2.99)^{3}-4\times2.99 = 26.730899 - 11.96 = 14.770899 \).

Step2: Compute the difference quotient

Substitute into \( \frac{f(3 + h)-f(3)}{h} \) (here \( h=-0.01 \)):
\( \frac{14.770899 - 15}{-0.01}=\frac{-0.229101}{-0.01}=22.9101 \).

For \( h = -0.1 \):

Answer:

s:

• \( h = 0.1 \): \( 23.91 \)
• \( h = 0.01 \): \( 23.0901 \)
• \( h = -0.01 \): \( 22.9101 \)
• \( h = -0.1 \): \( 22.11 \)
• Derivative at \( x = 3 \): \( 23 \)