I am trying to maximize an exponential function

Question

I am looking for the value of $x$ that will maximize $y$ in the following equation

$$ y=e^{-(x-a)^2/b} $$

where $a$ and $b$ are constants.

Any help is appreciated

Answer 1

Hint: Find local maxima using differentiation. Then check if any of these is a global maximum by taking limits at infinities considering whether the function is continuous and sketching the graph.

Answer 2

You can use the derivative test, $f'(x)=-2\frac{x-a}{b}f(x)$. Now since $f(x) >0$ then then we have only a critical point at $x=a$ and its nature might depend on the sign of b.

Answer 3

First, I assume that $b>0$.

Since $y>0$, and since $\ln$ is an increasing and one-to-one function, when $y$ is maximized, $\ln y$ is also maximized and vice versa. Therefore take $\ln$ and maximize, i.e. \begin{align} f(x)&=\ln e^{-\frac{(x-a)^2}{b}}\\ &=-\frac{(x-a)^2}{b} \end{align}

now since we assumed that $b>0$ we have that $-\frac{(x-a)^2}{b}\leq0$ whose max (which is $0$) happens when $x=a$.

If $b<0$ then $-\frac{(x-a)^2}{b}\geq0$ and its maximum is $+\infty$.

Answer 4

The given standard Gaussian distribution Bell curve has a maximum value at x = a as:

$$ y_\max =1 , b >0$$

btw, it is not the exponential function $ y = y_0 e^{\pm x/b} $ without maximum.

Answer 5

Since $w\mapsto e^w$ is an increasing function, the value that maximizes $e^w$ is the value that maximizes $w=-(x-a)^2/b$.

You haven't told us whether $b$ is positive or negative. If $b$ is positive, then you need the value of $x$ that minimizes $(x-a)^2$.