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
(a) the derivative of a function $f$ at $x$, denoted by $f^{prime}(x)$, is $f^{prime}(x)=lim _{h
ightarrow 0} \frac{f(x+h)-f(x)}{h}$ provided that the limit exists. use the
of the derivative of $f$ at $x$ to find the derivative of the given function $f(x)$ when $x = 1$

$$f^{prime}(1)=lim _{h ightarrow 0} \frac{left(1+h)^{2}+2 ight-left(1)^{2}+2 ight}{h}$$

evaluate the limit expression to find $f^{prime}(1)$.
$f^{prime}(1)=square$ (type an integer or a fraction.)

Explanation:

Step1: Expand the numerator

$$\begin{align*} &\frac{[(1 + h)^2+2]-[(1)^2 + 2]}{h}\\ =&\frac{(1 + 2h+h^2+2)-(1 + 2)}{h}\\ =&\frac{1 + 2h+h^2+2 - 3}{h}\\ =&\frac{2h+h^2}{h} \end{align*}$$

Step2: Simplify the fraction

$$ \frac{2h+h^2}{h}=\frac{h(2 + h)}{h}=2 + h $$

Step3: Evaluate the limit

$$ \lim_{h ightarrow0}(2 + h)=2+0 = 2 $$

Answer:

\(2\)