5. (4 pts) Use the limit definition of the derivative to find the derivative function of \( f(x)=x^{3}-5 x+1 \) note: \( (x+h)^{3}=x^{3}+3 x^{2} h+3 x h^{2}+h^{3} \)

To find the derivative of the function \( f(x) = x^3 - 5x + 1 \) using the limit definition of the derivative, we start with the formula: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] First, we calculate \( f(x+h) \): \[ f(x+h) = (x+h)^3 - 5(x+h) + 1 \] Substituting the binomial expansion \( (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \): \[ f(x+h) = (x^3 + 3x^2h + 3xh^2 + h^3) - 5x - 5h + 1 \] Now, simplify \( f(x+h) \): \[ f(x+h) = x^3 + 3x^2h + 3xh^2 + h^3 - 5x - 5h + 1 \] Next, we compute \( f(x+h) - f(x) \): \[ f(x) = x^3 - 5x + 1 \] Thus, \[ f(x+h) - f(x) = (x^3 + 3x^2h + 3xh^2 + h^3 - 5x - 5h + 1) - (x^3 - 5x + 1) \] This simplifies to: \[ f(x+h) - f(x) = 3x^2h + 3xh^2 + h^3 - 5h \] Now, we substitute this into the limit definition: \[ f'(x) = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3 - 5h}{h} \] Factor out \( h \): \[ f'(x) = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2 - 5)}{h} \] Canceling \( h \) from the numerator and denominator (assuming \( h \neq 0 \)), we get: \[ f'(x) = \lim_{h \to 0} (3x^2 + 3xh + h^2 - 5) \] Now as \( h \to 0 \): \[ f'(x) = 3x^2 - 5 \] Thus, the derivative function of \( f(x) = x^3 - 5x + 1 \) is: \[ f'(x) = 3x^2 - 5 \]