Question

If $y=alogx+b{x}^2+x$ has extreme values at $x=-1$ and $x=2$. Find a and b.

Answer 1

At extreme values, $\frac{dy}{dx}$ at that point is zero

$y=alogx+b{x}^{2}x$

x = - 1 is an extreme value

$(\frac{dy}{dx}{)}_{x=-1}=0$

$\frac{dy}{dx}=\frac{9}{x}+2bx+1$

Now , $(\frac{dy}{dx}{)}_{x=-1},\frac{+9}{-1}+2b(1)+1$

$\therefore -a-2b+1=0$

$\therefore a+2b=1$ -- (1)

x = 2 is an extreme value

$\therefore (\frac{dy}{dx}{)}_{x=2}$

$\therefore (\frac{dy}{dx}{)}_{x=2}=\frac{9}{2}+2b(2)+1$

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

$\therefore a+8b=-1$ _ (2)

subtracting equation (2) by (1), we get

6b = -2

$b=\frac{-1}{3}$

substituting the above value in equation (1)

$\therefore a+2\left(\frac{-1}{3}\right)=1$

$\therefore a=\frac{5}{3}$