Question

Find the extreme value of Quadratic expression $2x-7-5x^2$. Also state whether it is maximum or minimum with reason.

Answer 1

Let

$f\left(x\right)=-7+2x-5x^2$

$f^{\prime }\left(x\right)=2-10x$

Now $f^{\prime }\left(x\right)=0$ implies $2-10x=0$ or $x=\frac{1}{5}$.

Now

$f"\left(x\right)=-10$.

Since $f"\left(x\right)<0$ for all $x\in R$, hence $f\left(x\right)$ attains a maximum value at $x=\frac{1}{5}$.

Thus the maximum value is

$f\left(\frac{1}{5}\right)=-7+\frac{2}{5}-5\left(\frac{1}{25}\right)$

$=-7+\frac{2}{5}-\frac{1}{5}$

$=\frac{1}{5}-7$

$=\frac{-34}{5}$.