Question

Find the following for the given function f:

1. $f\text{'}\left(a\right),$where ais in the domain of f, and
2. $f\text{'}\left(3\right)$and$f\text{'}\left(4\right)$.

.$f\left(x\right)=x^2+2x$

Short answer

(a) The value of$f\text{'}\left(a\right)=2a+2$

(b) The value of $f\text{'}\left(3\right)=8andf\text{'}\left(4\right)=10$.

Step-by-step solution

Step 1. Given information.

The function here given is,

$f\left(x\right)=x^2+2x$

Step 2. Formula used.

The derivative of a function f at a number,denoted by$f\text{'}\left(a\right),$is

.$f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f(a+h)-f\left(a\right)}{h}$

if this limit exists.

Step 3. Finding the slope of the tangent line at a given point.

According to the definition of a derivative at a, we have

$\begin{array}{c}f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f(a+h)-f\left(a\right)}{h}Definitionoff\text{'}\left(4\right)\\ =\underset{h\to 0}{\lim }\frac{[{(a+h)}^2+2(a+h\left)\right]-[a^2+2a]}{h}f\left(x\right)=x^2+2x\\ =\underset{h\to 0}{\lim }\frac{a^2+h^2+2ah+2a+2h-a^2-2a}{h}Expand\\ =\underset{h\to 0}{\lim }\frac{h^2+2ah+2h}{h}Simplify\\ =\underset{h\to 0}{\lim }(h+2a+2)Cancelh\\ =2a+2\end{array}$

Therefore, the derivative of the function at a is $f\text{'}\left(a\right)=2a+2$.

Step 4. Finding the value of derivative at given point.

We have.$f\text{'}\left(a\right)=2a+2$

Let’s find the value of derivative at$a=3anda=4$$a=3anda=4$.

Substitute the value of$a=3anda=4$ into the$f\text{'}\left(a\right)=2a+2$

$\begin{array}{c}f\text{'}\left(3\right)=2\left(3\right)+2\\ =6+2\\ =8\\ f\text{'}\left(4\right)=2\left(4\right)+2\\ =8+2\\ =10\end{array}$

Therefore, the value of$f\text{'}\left(3\right)=8andf\text{'}\left(4\right)=10$ .