Question

using the definition, calculate the derivative of the function. then find the values of the derivative as specified.
$f(x)=7 - x^{2};f(-6),f(0),f(5)$

Explanation:

Step1: Recall the definition of the derivative

The derivative of a function \(y = f(x)\) is defined as \(f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}\).
Given \(f(x)=7 - x^{2}\), then \(f(x + h)=7-(x + h)^{2}=7-(x^{2}+2xh+h^{2})=7 - x^{2}-2xh - h^{2}\).
So \(\frac{f(x + h)-f(x)}{h}=\frac{(7 - x^{2}-2xh - h^{2})-(7 - x^{2})}{h}\).

Step2: Simplify the expression

$$
LATEXBLOCK0
$$

Step3: Find the limit as \(h

ightarrow0\)
\(f^{\prime}(x)=\lim_{h
ightarrow0}(-2x - h)\).
As \(h
ightarrow0\), we have \(f^{\prime}(x)=-2x\).

Step4: Calculate \(f^{\prime}(-6)\), \(f^{\prime}(0)\) and \(f^{\prime}(5)\)

When \(x=-6\), \(f^{\prime}(-6)=-2\times(-6) = 12\).
When \(x = 0\), \(f^{\prime}(0)=-2\times0=0\).
When \(x = 5\), \(f^{\prime}(5)=-2\times5=-10\).

Answer:

\(f^{\prime}(x)=-2x\), \(f^{\prime}(-6) = 12\), \(f^{\prime}(0)=0\), \(f^{\prime}(5)=-10\)