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 $f(x)=limlimits_{h\to0}\frac{f(x + h)-f(x)}{h}$ 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 $x = 1$.
$f(1)=limlimits_{h\to0}\frac{(1 + h)^{2}-4-(1)^{2}-4}{h}$
evaluate the limit expression to find $f(1)$.
$f(1)=2$ (type an integer or a fraction.)
(b) $f(1)=-3$ (type an integer or a fraction)
the equation of the tangent line at $(1, f(1))$ is $2x - 5$. (type your answer in slope - intercept form.)

Explanation:

Step1: Find \(f(1)\)

Substitute \(x = 1\) into \(f(x)=x^{2}-4\).
\(f(1)=1^{2}-4=-3\)

Step2: Find the derivative \(f^{\prime}(1)\)

$$
LATEXBLOCK0
$$

Step3: Find the equation of the tangent line

The slope - intercept form of a line is \(y=mx + b\), where \(m\) is the slope and \(b\) is the \(y\) - intercept.
We know that \(m = f^{\prime}(1)=2\) and the line passes through the point \((1,f(1))=(1,-3)\).
Substitute \(x = 1\), \(y=-3\) and \(m = 2\) into \(y=mx + b\)
\(-3=2\times1+b\)
\(b=-3 - 2=-5\)
The equation of the tangent line is \(y = 2x-5\)

Answer:

(a) \(f^{\prime}(1)=2\)
(b) \(f(1)=-3\), equation of the tangent line \(y = 2x-5\)