Let's analyze the given function step by step.
### Part a: Difference Quotient
Given the function [tex]\( f(x) = -2x^2 + 5x \)[/tex].
The difference quotient is given by [tex]\( \frac{f(x + h) - f(x)}{h} \)[/tex].
We need to find [tex]\( f(x + h) \)[/tex]:
[tex]\[ f(x + h) = -2(x + h)^2 + 5(x + h) \][/tex]
First, expand [tex]\( (x + h)^2 \)[/tex]:
[tex]\[ (x + h)^2 = x^2 + 2xh + h^2 \][/tex]
Now, substitute back:
[tex]\[ f(x + h) = -2(x^2 + 2xh + h^2) + 5(x + h) \][/tex]
[tex]\[ f(x + h) = -2x^2 - 4xh - 2h^2 + 5x + 5h \][/tex]
Next, find [tex]\( f(x + h) - f(x) \)[/tex]:
[tex]\[ f(x + h) - f(x) = (-2x^2 - 4xh - 2h^2 + 5x + 5h) - (-2x^2 + 5x) \][/tex]
[tex]\[ f(x + h) - f(x) = -2x^2 - 4xh - 2h^2 + 5x + 5h + 2x^2 - 5x \][/tex]
[tex]\[ f(x + h) - f(x) = -4xh - 2h^2 + 5h \][/tex]
Now, simplify [tex]\(\frac{f(x + h) - f(x)}{h} \)[/tex]:
[tex]\[ \frac{f(x + h) - f(x)}{h} = \frac{-4xh - 2h^2 + 5h}{h} \][/tex]
[tex]\[ \frac{f(x + h) - f(x)}{h} = -4x - 2h + 5 \][/tex]
So, the simplified difference quotient is:
[tex]\[ \frac{f(x+h)-f(x)}{h} = -4x - 2h + 5 \][/tex]
### Part b: Derivative
The derivative of the function [tex]\( f'(x) \)[/tex] is the limit of the difference quotient as [tex]\( h \)[/tex] approaches zero:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \][/tex]
Given the simplified difference quotient:
[tex]\[ f'(x) = \lim_{h \to 0} (-4x - 2h + 5) \][/tex]
Taking the limit as [tex]\( h \)[/tex] approaches zero:
[tex]\[ f'(x) = -4x + 5 \][/tex]
So, the derivative [tex]\( f'(x) \)[/tex] is:
[tex]\[ f^{\prime}(x) = -4x + 5 \][/tex]
Therefore, the final answers are:
a. The simplified difference quotient is:
[tex]\[ \frac{f(x + h) - f(x)}{h} = -4x - 2h + 5 \][/tex]
b. The derivative of the function is:
[tex]\[ f^{\prime}(x) = -4x + 5 \][/tex]
