Question

for the function ( f(x) = -2x^2 ): (a) a simplified form of the difference quotient ( \frac{f(x+h) - f(x)}{h} ), when ( h
eq 0 ),

Explanation:

Step1: Find \( f(x + h) \)

Substitute \( x + h \) into \( f(x) = -2x^2 \). So \( f(x + h) = -2(x + h)^2 \). Expand \( (x + h)^2 \) using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \), we get \( f(x + h) = -2(x^2 + 2xh + h^2) = -2x^2 - 4xh - 2h^2 \).

Step2: Compute \( f(x + h) - f(x) \)

Substitute \( f(x + h) = -2x^2 - 4xh - 2h^2 \) and \( f(x) = -2x^2 \) into \( f(x + h) - f(x) \). So \( f(x + h) - f(x) = (-2x^2 - 4xh - 2h^2) - (-2x^2) \). Simplify this: \( -2x^2 - 4xh - 2h^2 + 2x^2 = -4xh - 2h^2 \).

Step3: Divide by \( h \) (since \( h

eq 0 \))
Now, divide \( f(x + h) - f(x) = -4xh - 2h^2 \) by \( h \). So \( \frac{f(x + h) - f(x)}{h} = \frac{-4xh - 2h^2}{h} \). Factor out \( h \) from the numerator: \( \frac{h(-4x - 2h)}{h} \). Cancel out \( h \) (since \( h
eq 0 \)), we get \( -4x - 2h \).

Answer:

\( -4x - 2h \)