Question

If f x is equals to a log x + b x square + extreme value at x is equals to minus one and x is equals to 2 then find the value of a and b

Answer 1

Since, $f(x)= a logx +bx^{2}+x$

Since y = f(x), therefore we can write the function as:

$y= a logx +bx^{2}+x$

Since the above function has extreme values at x = -1 and x =2.

Let us find the derivative of the function f(x), we get

$y'(x)= \frac{a}{x} +2bx+1$

Since -1 and 2 are the extreme values, therefore

y'(-1)= 0 and y'(2)=0

So, Consider y'(-1)=0

$y'= \frac{a}{-1} +2b(-1)+1=0$

$-a -2b=-1$

So, a+2b=1 (Equation 1)

Therefore, a = 1-2b

Consider y'(2)= 0

$y'= \frac{a}{2}+2b(2)+1=0$

So, $\frac{a}{2}+4b=-1$

$a+8b= -2$   (Equation 2)

Solving equation 1 and 2, we get

Putting the value of a in equation 2, we get

a+8b= -2

1-2b+8b = -2

1+6b = -2

6b = -3

$b = \frac{-1}{2}$

Since, a = 1-2b

$a=1-2(\frac{-1}{2}) = 2$

Therefore, the value of 'a' is 2 and 'b' is $\frac{-1}{2}$.