Question

use the limit definition of the derivative to answer the following questions about $f(x) = 3x^2$. find the difference quotient $\frac{f(x + h) - f(x)}{h}$ \boxed{} find $f(x)$ by determining $\lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h}$. \boxed{} find $f(-2)$ \boxed{} find $f(0)$ \boxed{} find $f(1)$ \boxed{}

Explanation:

Part 1: Find the difference quotient $\frac{f(x + h) - f(x)}{h}$

Step 1: Compute $f(x + h)$

Given $f(x) = 3x^2$, substitute $x + h$ into the function:
$f(x + h) = 3(x + h)^2$
Expand $(x + h)^2$ using the formula $(a + b)^2 = a^2 + 2ab + b^2$:
$f(x + h) = 3(x^2 + 2xh + h^2) = 3x^2 + 6xh + 3h^2$

Step 2: Compute $f(x + h) - f(x)$

Subtract $f(x) = 3x^2$ from $f(x + h)$:
$f(x + h) - f(x) = (3x^2 + 6xh + 3h^2) - 3x^2$
Simplify by canceling $3x^2$:
$f(x + h) - f(x) = 6xh + 3h^2$

Step 3: Divide by $h$ (and simplify)

Divide the result by $h$ (assuming $h
eq 0$):
$\frac{f(x + h) - f(x)}{h} = \frac{6xh + 3h^2}{h}$
Factor out $h$ in the numerator:
$\frac{h(6x + 3h)}{h}$
Cancel $h$:
$\frac{f(x + h) - f(x)}{h} = 6x + 3h$

Part 2: Find $f'(x)$ by determining $\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Step 1: Use the difference quotient from Part 1

We know $\frac{f(x + h) - f(x)}{h} = 6x + 3h$. Now take the limit as $h \to 0$:
$\lim_{h \to 0} (6x + 3h)$

Step 2: Evaluate the limit

As $h \to 0$, the term $3h$ approaches $0$. Thus:
$\lim_{h \to 0} (6x + 3h) = 6x + 3(0) = 6x$
So, $f'(x) = 6x$

Part 3: Find $f'(-2)$

Step 1: Use the derivative formula from Part 2

We found $f'(x) = 6x$. Substitute $x = -2$:
$f'(-2) = 6(-2)$

Step 2: Simplify

$6(-2) = -12$

Part 4: Find $f'(0)$

Answer:

s:

• Difference quotient: $\boldsymbol{6x + 3h}$
• $f'(x)$: $\boldsymbol{6x}$
• $f'(-2)$: $\boldsymbol{-12}$
• $f'(0)$: $\boldsymbol{0}$
• $f'(1)$: $\boldsymbol{6}$