Question

For what values of the number \(a\) and \(b\) does the function \(f(x) = ax{e^{b{x^2}}}\) have the maximum value \(f(2) = 1\)?

Short answer

The resultant answer is \(a = \frac{{\sqrt e }}{2}\) and \(b = - \frac{1}{8}\).

Step-by-step solution

Given data

The given function is \(f(x) = ax{e^{b{x^2}}}\).

Concept of Differentiation

Differentiation is a method of finding the derivative of a function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables.

Determine the derivative of the function

The expression is: \(f(2) = a(2){e^{b{{(2)}^2}}} = 2a{e^{4b}} = 1\)

The derivative tests:

\(\begin{aligned}{c}{f^\prime }(x) = a{e^{b{x^2}}} + ax(2bx){e^{b{x^2}}} = 0\\x{e^{b{x^2}}}\left( {1 + 2b{x^2}} \right) = 0\end{aligned}\)

Plug 2 in for \(x\).

\(\begin{aligned}{c}a{e^{b{x^2}}} = 0{\rm{ or }}1 + 8b = 0\\a = 0{\rm{ or }}b = - \frac{1}{8}\end{aligned}\)

But if \(a = 0\), then \(f(x)\)would be always zero.

\(\begin{aligned}{c}{f^{\prime \prime }}(x) &= a(2bx){e^{b{x^2}}} + 4abx{e^{b{x^2}}} + 2ab{x^2}(2bx){e^{b{x^2}}}\\{f^{\prime \prime }}(x) &= 2abx{e^{b{x^2}}} + 4abx{e^{b{x^2}}} + 4a{b^2}{x^3}{e^{b{x^2}}}\\{f^{\prime \prime }}(x) &= 2abx{e^{b{x^2}}}\left( {3 + 2b{x^2}} \right)\\{f^{\prime \prime }}(x) &= 0\end{aligned}\)

Simplify the expression

Plug 2 in for \(x\); \(4ab{e^{4b}}(3 + 8b) < 0\)

For the product to be less than zero (so the function is concave down giving a max), the factors have to have opposite signs. \(4ab{e^{4b}} > 0{\rm{ and }}3 + 8b < 0\), \(ab > 0{\rm{ and }}b < - \frac{3}{8}\) but \(b = - \frac{1}{8} > - \frac{3}{8}{\rm{ so }}a < 0{\rm{ and }}b > - \frac{3}{8}\).

Calculate the value of \(a\):

\(\begin{aligned}{c}f(2) = a(2){e^{b{{(2)}^2}}} = 2a{e^{4b}} &= 1\\2a{e^{ - \frac{1}{2}}} &= 1\\\frac{{2a}}{{\sqrt e }} &= 1\\a &= \frac{{\sqrt e }}{2}\end{aligned}\)

The resultant answer is \(a = \frac{{\sqrt e }}{2}\) and \(b = - \frac{1}{8}\).