Question

use the function below to answer parts (a)-(c).

$f(x)=x^{2}-4$

(a) use the formal definition to find the derivative of $y = f(x)$ at $x = 1$.
(b) find $f(1)$ and find the equation of the tangent line at the point $(1,f(1))$
(c) graph $y = f(x)$ and the tangent line at the point $(1,f(1))$ in the same coordinate system.

(a) the derivative of a function $f$ at $x$, denoted by $f(x)$, is $f(x)=limlimits_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$ provided that the limit exists. use the
finition of the derivative of $f$ at $x$ to find the derivative of the given function $f(x)$ when $x = 1$.

$f(1)=limlimits_{h
ightarrow0}\frac{(1 + h)^{2}-4-(1)^{2}-4}{h}$

evaluate the limit expression to find $f(1)$.

$f(1)=\square$ (type an integer or a fraction.)

Explanation:

Step1: Expand the numerator

First, expand \((1 + h)^{2}-4\) and \((1)^{2}-4\).
\((1 + h)^{2}-4=1 + 2h+h^{2}-4=h^{2}+2h - 3\)
\((1)^{2}-4=1 - 4=-3\)
So the numerator \([(1 + h)^{2}-4]-[(1)^{2}-4]=(h^{2}+2h - 3)-(-3)=h^{2}+2h\)

Step2: Simplify the limit expression

The limit expression \(f^{\prime}(1)=\lim_{h
ightarrow0}\frac{[(1 + h)^{2}-4]-[(1)^{2}-4]}{h}=\lim_{h
ightarrow0}\frac{h^{2}+2h}{h}\)
Since \(h
eq0\) (as \(h
ightarrow0\) but \(h
eq0\) in the limit process), we can cancel out the \(h\) terms. \(\frac{h^{2}+2h}{h}=h + 2\)

Step3: Evaluate the limit

Now, find \(\lim_{h
ightarrow0}(h + 2)\). Substitute \(h = 0\) into \(h+2\). \(\lim_{h
ightarrow0}(h + 2)=0 + 2\)

Answer:

\(2\)