Question

In Problems$33–35$, find the derivative of each function at the number indicated
$\text{34.}f\left(x\right)=x^2-3x$at$0$

Short answer

The derivative of $f\left(x\right)=x^2-3x$at $0$ is$f\text{'}\left(0\right)=-3$.

Step-by-step solution

Step 1. Find f(0)

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 $f\left(x\right)=x^2-3x$and $0$

Now we have to find, therefore substitute $c=0$in $f\left(x\right)$to find$f\left(0\right)$.

$\begin{array}{rl}f\left(x\right)& =x^2-3x\\ f\left(0\right)& =(0{)}^2-3(0)\\ & =0\end{array}$

Step 2. Use the derivative formula

Substituting all the values in the derivative formula we get,

$\begin{array}{rl}{f}^{\mathrm{\prime }}\left(0\right)& =\underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-\left(0\right)}\\ & =\underset{x\to 0}{lim}\frac{x^2-3x-0}{x}\\ & =\underset{x\to 0}{lim}\frac{x^2-3x}{x}\\ & =\underset{x\to 0}{lim}\frac{x(x-3)}{x}\end{array}$

$\begin{array}{rl}& =\underset{x\to 0}{lim}(x-3)\\ & =0-3\\ & =-3\end{array}$

Hence the derivative of $f\left(x\right)$when$c=0$is$f\text{'}\left(0\right)=-3$.