Question

attempt 1: 10 attempts remaining. let ( f(x) = 2x^2 - 2x - 7 ). (a) compute the difference quotient of ( f(x) ). ( \frac{f(x + h) - f(x)}{h} = ) (b) compute the derivative of ( f(x) ). ( f(x) = ) submit answer next item

Explanation:

Part (a)

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

Given \( f(x) = 2x^2 - 2x - 7 \), substitute \( x + h \) into \( f \):
\( f(x + h) = 2(x + h)^2 - 2(x + h) - 7 \)
Expand \( (x + h)^2 \): \( (x + h)^2 = x^2 + 2xh + h^2 \)
So, \( f(x + h) = 2(x^2 + 2xh + h^2) - 2x - 2h - 7 = 2x^2 + 4xh + 2h^2 - 2x - 2h - 7 \)

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

Subtract \( f(x) = 2x^2 - 2x - 7 \) from \( f(x + h) \):
\( f(x + h) - f(x) = (2x^2 + 4xh + 2h^2 - 2x - 2h - 7) - (2x^2 - 2x - 7) \)
Simplify by canceling \( 2x^2 \), \( -2x \), and \( -7 \):
\( = 4xh + 2h^2 - 2h \)

Step 3: Divide by \( h \) ( \( h

eq 0 \))
\( \frac{f(x + h) - f(x)}{h} = \frac{4xh + 2h^2 - 2h}{h} \)
Factor out \( h \) from the numerator:
\( = \frac{h(4x + 2h - 2)}{h} \)
Cancel \( h \):
\( = 4x + 2h - 2 \)

Step 1: Recall the definition of the derivative

The derivative \( f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \)
From part (a), we have \( \frac{f(x + h) - f(x)}{h} = 4x + 2h - 2 \)

Step 2: Take the limit as \( h \to 0 \)

\( f'(x) = \lim_{h \to 0} (4x + 2h - 2) \)
As \( h \to 0 \), \( 2h \to 0 \), so:
\( f'(x) = 4x - 2 \)

Answer:

(a):
\( 4x + 2h - 2 \)

Part (b)