Question

Find constants a and b in the function f(x)=axe^bx such that f(1/5)=1 and the function has a local maximum at x=1/5.

Answer 1

F (x) = axe ^ bx

F ‘ (x) = d/dx (axe ^ bx)

= a (d/dx) (xe ^ bx)

= a [(d/dx * x) e ^ bx + (x * d/dx * e ^ bx)]

= a [e ^ bx + bxe ^ bx]

= a [1 + bx] e ^ bx

 

For critical points, set f ‘ (x) = 0 and solve for x

 

a [1 + bx] e ^ bx = 0

= 1 + bx = 0

= bx = -1

X = -1/b

Given that f (x) has local max at 1/5, critical point x = 1/5

-1/5 = -1/b

Therefore b = -5

 

F(x) = axe ^ -5x

F(1/5) = a/5 e ^-1

1 = a/5e

 

Therefore, a = 5e

Answer 2

Answer:

The value of a and b is $5e$ and $-5$, respectively.

Step-by-step explanation:

The given function is $f(x)=axe^{bx}$.

The value of function at $x=\dfrac{1}{5}$ is $f(\dfrac{1}{5})=1$ and the function has a local maxima at  $x=\dfrac{1}{5}$.

So, the first derivative of the function at $x=\dfrac{1}{5}$ will be zero.

Now, calculating for a and b.

$f(x)=axe^{bx}\\f(\dfrac{1}{5})=a\left (\dfrac{1}{5}\right )e^{b\dfrac{1}{5}}\\1=\left (\dfrac{a}{5}\right )e^{\dfrac{b}{5}}$

Differentiate the function,

$f(x)=axe^{bx}\\f'(x)=abxe^{bx}+ae^{bx}\\f'(1/5)=\dfrac{ab}{5}e^{\dfrac{b}{5}}+ae^{\dfrac{b}{5}}=0\\b=-5$

Solving for a as,

$1=\left (\dfrac{a}{5}\right )e^{\dfrac{b}{5}}\\a=5e$

Therefore, the value of a and b is $5e$ and $-5$, respectively.

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