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