Question

If $y=a\log x+b{x}^2+x$ has its extreme value at $x=-1$ and $x=2$, then find $a$ and $b$.

Answer 1

Let, $y=a\log x+b{x}^2+x$ has its extreme value at $x=-1$ and $x=2$.

Then, ${\frac{dy}{dx}|}_{x=-1}$ $=0$ and ${\frac{dy}{dx}|}_{x=2}$ $=0$.

Now $\frac{dy}{dx}=\frac{a}{x}$ $+2bx+1$.

For, ${\frac{dy}{dx}|}_{x=-1}$ $=0$ we have

$-a-2b+1=0$

or, $a+2b=1$......(1).

Also, ${\frac{dy}{dx}|}_{x=2}$ $=0$ gives

$\frac{a}{2}$ $+4b+1=0$

or, $a+8b=-2$.......(2).

Now, subtracting (1) from (2) we get,

$6b=-3\Rightarrow b=-\frac{1}{2}$.

Now from (1) using the value of $b$ we get, $a=2$.