Question

Find the derivative of the $f(x)=2x^2+3x-5$ at $x=-1$

• A. $0$
• B. $1$
• C. $-1$
• D. $-2$

Answer 1

Answer: (C)

We have, $f\left(x\right)=2x^2+3x-5$
$\therefore f^{\prime }\left(-1\right)$ = $\underset{h\to 0}{\lim }$$\frac{f\left(-1+h\right)-f\left(-1\right)}{h}$
$=\underset{h\to 0}{\lim }$ $\frac{\left[2\left(-1+h^2\right)+3\left(-1+h\right)-5\right]-\left[2{\left(-1\right)}^2+3\left(-1\right)-5\right]}{h}$
$=\underset{h\to 0}{\lim }$$\frac{2h^2-h}{h}$
$=\underset{h\to 0}{\lim }$$\left(2h-1\right)$
$=2\left(0\right)-1=-1$