Question

introduction to derivatives: problem 1 (1 point) let ( f(x) = x^2 - 15x ). 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? note: you can earn partial credit on this problem. preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining. email instructor page generated september 6, 2025 at 4:43:35 pm cdt webwork © 1996-2024 | theme: math4 | ww_version: 2.19-10 | pg_version the webwork project

Explanation:

First, we find \( f(3 + h) \) and \( f(3) \) for the function \( f(x)=x^{2}-15x \).

Step 1: Find \( f(3 + h) \)

Substitute \( x = 3+h \) into \( f(x) \):
\[

$$\begin{align*} f(3 + h)&=(3 + h)^{2}-15(3 + h)\\ &=9 + 6h+h^{2}-45 - 15h\\ &=h^{2}-9h - 36 \end{align*}$$

\]

Step 2: Find \( f(3) \)

Substitute \( x = 3 \) into \( f(x) \):
\[
f(3)=3^{2}-15\times3=9 - 45=- 36
\]

Step 3: Compute the difference quotient \( \frac{f(3 + h)-f(3)}{h} \)

Substitute \( f(3 + h) \) and \( f(3) \) into the difference quotient:
\[

$$\begin{align*} \frac{f(3 + h)-f(3)}{h}&=\frac{(h^{2}-9h - 36)-(-36)}{h}\\ &=\frac{h^{2}-9h - 36 + 36}{h}\\ &=\frac{h^{2}-9h}{h}\\ &=h - 9\quad(h eq0) \end{align*}$$

\]

Now we calculate the difference quotient for different values of \( h \):

For \( h = 0.1 \)

Substitute \( h = 0.1 \) into \( h - 9 \):
\[
0.1-9=-8.9
\]

For \( h = 0.01 \)

Substitute \( h = 0.01 \) into \( h - 9 \):
\[
0.01-9=-8.99
\]

For \( h=- 0.01 \)

Substitute \( h = - 0.01 \) into \( h - 9 \):
\[
-0.01-9=-9.01
\]

For \( h=-0.1 \)

Substitute \( h=-0.1 \) into \( h - 9 \):
\[
-0.1-9=-9.1
\]

Step 4: Find the derivative at \( x = 3 \)

The derivative of \( f(x) \) at \( x = 3 \) is the limit of the difference quotient as \( h\to0 \). From the difference quotient \( \frac{f(3 + h)-f(3)}{h}=h - 9 \), as \( h\to0 \), the limit is \( 0-9=-9 \).

For \( h = 0.1 \):

We use the simplified difference quotient \( h - 9 \). Substitute \( h = 0.1 \).

Step 1: Substitute \( h = 0.1 \) into \( h - 9 \)

\( 0.1-9=-8.9 \)

Use the difference quotient \( h - 9 \), substitute \( h = 0.01 \).

Step 1: Substitute \( h = 0.01 \) into \( h - 9 \)

\( 0.01-9=-8.99 \)

Use the difference quotient \( h - 9 \), substitute \( h=-0.01 \).

Step 1: Substitute \( h = - 0.01 \) into \( h - 9 \)

\( -0.01-9=-9.01 \)

Answer:

\( -8.9 \)

For \( h = 0.01 \):