Question

Find the derivative of the function at the given number.

.$f\left(x\right)=2-3x+x^2,at-1$

Short answer

The derivative of the function at -1is-5.

Step-by-step solution

Step 1. Given information.

The function here given is,

$f\left(x\right)=2-3x+x^2,at-1$

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, with$a=-1$, we have

$\begin{array}{c}f\text{'}(-1)=\underset{h\to 0}{\lim }\frac{f(-1+h)-f(-1)}{h}Definitionoff\text{'}(-1)\\ =\underset{h\to 0}{\lim }\frac{[2-3(-1+h)+{(-1+h)}^2]-[2-3(-1)+{(-1)}^2]}{h}f\left(x\right)=2-3x+x^2\\ =\underset{h\to 0}{\lim }\frac{[2+3-3h+(1+h^2-2h\left)\right]-6}{h}Expand\\ =\underset{h\to 0}{\lim }\frac{6-5h+h^2-6}{h}\\ =\underset{h\to 0}{\lim }\frac{h^2-5h}{h}Simplify\\ =\underset{h\to 0}{\lim }(h-5)Cancelh\\ =-5\end{array}$

Therefore, the derivative of the function at is-1 .-5