Questions: Consider the function (f(x)=5 x-x^2) Use the limit definition to compute the following derivative values: (f^prime(0)=) (f^prime(1)=) (f^prime(2)=) (f^prime(3)=)

Solution Steps

To find the derivative of the function $f(x) = 5x - x^2$ at specific points using the limit definition, we use the formula for the derivative: $$f'(a) = \lim_{{h \to 0}} \frac{f(a+h) - f(a)}{h}$$

1. Substitute $f(x) = 5x - x^2$ into the limit definition.
2. Compute the limit for each given point $a$.

Given the function $f(x) = 5x - x^2$, we use the limit definition of the derivative: $$f'(a) = \lim_{{h \to 0}} \frac{f(a+h) - f(a)}{h}$$

Substitute $a = 0$ into the derivative formula: $$f'(0) = \lim_{{h \to 0}} \frac{f(0+h) - f(0)}{h}$$ $$f'(0) = \lim_{{h \to 0}} \frac{(5(0+h) - (0+h)^2) - (5(0) - 0^2)}{h}$$ $$f'(0) = \lim_{{h \to 0}} \frac{5h - h^2}{h}$$ $$f'(0) = \lim_{{h \to 0}} (5 - h)$$ As $h \to 0$, $f'(0) = 5$.

Substitute $a = 1$ into the derivative formula: $$f'(1) = \lim_{{h \to 0}} \frac{f(1+h) - f(1)}{h}$$ $$f'(1) = \lim_{{h \to 0}} \frac{(5(1+h) - (1+h)^2) - (5(1) - 1^2)}{h}$$ $$f'(1) = \lim_{{h \to 0}} \frac{5 + 5h - 1 - 2h - h^2 - 4}{h}$$ $$f'(1) = \lim_{{h \to 0}} \frac{4 + 3h - h^2}{h}$$ $$f'(1) = \lim_{{h \to 0}} (3 - h)$$ As $h \to 0$, $f'(1) = 3$.

Substitute $a = 2$ into the derivative formula: $$f'(2) = \lim_{{h \to 0}} \frac{f(2+h) - f(2)}{h}$$ $$f'(2) = \lim_{{h \to 0}} \frac{(5(2+h) - (2+h)^2) - (5(2) - 2^2)}{h}$$ $$f'(2) = \lim_{{h \to 0}} \frac{10 + 5h - 4 - 4h - h^2 - 6}{h}$$ $$f'(2) = \lim_{{h \to 0}} \frac{6 + h - h^2}{h}$$ $$f'(2) = \lim_{{h \to 0}} (1 - h)$$ As $h \to 0$, $f'(2) = 1$.

Final Answer