Answer to Question #98746 in Calculus for pargol sanaat

Question

Find constants a and b in the function f(x)=axe^(bx)such that f(1/4)=1 and the function has a local maximum at x=1/4.

Answer 1

Given, $f(x) = axe^{bx}$

$f(1/4)=1$

$(a/4)e^{b/4}=1 ----(1)$

Local maximum at x=1/4, that means $f'(1/4)=0$

$f'(x) = abxe^{bx} + ae^{bx}$

$f'(1/4)= (ab/4)e^{b/4}+ae^{b/4}=0 --(2)$

Solving the 2nd equation,

$e^{b/4}(ab/4+a)=0$

Now, $e^{b/4}≠0$

So, $(ab/4 +a)=0$

Or, $a(b/4+1)=0$

Now from equation (1), value of 'a' cannot be zero.

So, $(b/4+1)=0$

Or, $b=-4$

Putting the value of 'b' in equation (1),

$(a/4) e^{-1}=1$

$a/4e=1$

Or, $a=4e$

So, $a=4e,b= -4$