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

Limits and Derivatives

Solution:

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