Question

use the function below to answer parts (a)-(c).
$f(x)=x^{2}-2$
(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.
(b) $f(1)=$ (type an integer or a fraction.)
ion of the tangent line at $(1,f(1))$ is $y = 2x + 1$. (type your answer in slope - intercept form.)
(c) choose the correct graph below.

Explanation:

Step1: Find \( f(1) \)

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

Step2: Find the derivative using the formal definition

The formal definition of the derivative is \( f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h} \).
For \( f(x)=x^{2}-2 \), \( f(x + h)=(x + h)^{2}-2=x^{2}+2xh+h^{2}-2 \).
Then \( \frac{f(x + h)-f(x)}{h}=\frac{(x^{2}+2xh+h^{2}-2)-(x^{2}-2)}{h}=\frac{2xh+h^{2}}{h}=2x + h \).
Taking the limit as \( h
ightarrow0 \), \( f^{\prime}(x)=\lim_{h
ightarrow0}(2x + h)=2x \).
At \( x = 1 \), \( f^{\prime}(1)=2\times1 = 2 \).

Step3: Find the equation of the tangent line

The point - slope form of a line is \( y - y_{1}=m(x - x_{1}) \), where \( (x_{1},y_{1})=(1,f(1))=(1,-1) \) and \( m = f^{\prime}(1)=2 \).
\( y-(-1)=2(x - 1) \)
\( y + 1=2x-2 \)
\( y=2x-3 \)

Answer:

\( f(1)=-1 \), equation of the tangent line \( y = 2x-3 \)