Question

In Problems$33–35$, find the derivative of each function at the number indicated.
$\text{35.}f\left(x\right)=2x^2+3x+2$at $1$

Short answer

The derivative of $f\left(x\right)=2x^2+3x+2$at 1 is $f\text{'}\left(1\right)=7$.

Step-by-step solution

Step 1. Find f(1)

To find the derivative we have to use the formula,

$f^{\mathrm{\prime }}\left(c\right)=\underset{x\to c}{lim}\frac{f\left(x\right)-f\left(c\right)}{x-c}$

We have been given $f\left(x\right)=2x^2+3x+2$and$c=1$

Now we have to find, therefore substitute $c=1$in $f\left(x\right)$to find $f\left(1\right)$.

$\begin{array}{rl}f\left(x\right)& =2x^2+3x+2\\ f\left(1\right)& =2(1{)}^2+3(1)+2\\ & =2+3+2\\ & =7\end{array}$

Step 2. Use the derivative formula

Substituting all the values in the derivative formula we get,

$$\begin{gathered} \begin{array}{rl}{f}^{\mathrm{\prime }}\left(1\right)& =\underset{x\to 1}{lim}\frac{f\left(x\right)-f\left(1\right)}{x-\left(1\right)}\\ & =\underset{x\to 1}{lim}\frac{2x^2+3x+2-7}{x-1}\\ & =\underset{x\to 1}{lim}\frac{2x^2+3x-5}{x-1}\\ & =\underset{x\to 1}{lim}\frac{(2x+5)(x-1)}{x-1}\end{array} \\ \begin{array}{rl}& =\underset{x\to 1}{lim}(2x+5)\\ & =2\left(1\right)+5\\ & =2+5\\ & =7\end{array} \end{gathered}$$

Hence the derivative of $f\left(x\right)$when $c=1$is $f\text{'}\left(1\right)=7$.