Using the definition, calculate t of the derivative as specified. \( f(x)=2-x^{2} ; f^{\prime}(-1), f^{\prime}(0), f^{\prime}(4) \)

We are given the function \[ f(x)=2-x^2 \] and we want to find the derivative at the points \(x = -1\), \(0\), and \(4\) using the definition of the derivative: \[ f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h} \] ### 1. Calculation of \( f'(-1) \) - Compute \( f(-1) \): \[ f(-1)=2-(-1)^2=2-1=1 \] - Compute \( f(-1+h) \): \[ f(-1+h)=2-(-1+h)^2 \] Expand the square: \[ (-1+h)^2 = h^2 - 2h + 1 \] Thus, \[ f(-1+h)=2-(h^2-2h+1)=2-h^2+2h-1=1+2h-h^2 \] - Form the difference quotient: \[ \frac{f(-1+h)-f(-1)}{h}=\frac{(1+2h-h^2)-1}{h}=\frac{2h-h^2}{h} \] Simplify by factoring \( h \): \[ \frac{h(2-h)}{h}=2-h \quad \text{(for \( h \neq 0\))} \] - Take the limit as \( h\to 0 \): \[ f'(-1)=\lim_{h\to 0}(2-h)=2 \] ### 2. Calculation of \( f'(0) \) - Compute \( f(0) \): \[ f(0)=2-0^2=2 \] - Compute \( f(0+h)=f(h) \): \[ f(h)=2-h^2 \] - Form the difference quotient: \[ \frac{f(0+h)-f(0)}{h}=\frac{(2-h^2)-2}{h}=\frac{-h^2}{h}=-h \] - Take the limit as \( h\to 0 \): \[ f'(0)=\lim_{h\to 0}(-h)=0 \] ### 3. Calculation of \( f'(4) \) - Compute \( f(4) \): \[ f(4)=2-4^2=2-16=-14 \] - Compute \( f(4+h) \): \[ f(4+h)=2-(4+h)^2 \] Expand the square: \[ (4+h)^2=16+8h+h^2 \] So, \[ f(4+h)=2-(16+8h+h^2)=2-16-8h-h^2=-14-8h-h^2 \] - Form the difference quotient: \[ \frac{f(4+h)-f(4)}{h}=\frac{(-14-8h-h^2)-(-14)}{h}=\frac{-8h-h^2}{h} \] Factor \( h \): \[ \frac{h(-8-h)}{h}=-8-h \quad \text{(for \( h \neq 0 \))} \] - Take the limit as \( h\to 0 \): \[ f'(4)=\lim_{h\to 0}(-8-h)=-8 \] ### Final Answers \[ f'(-1)=2, \quad f'(0)=0, \quad f'(4)=-8 \]