How do you find critical points of a logarithmic function $y = e^x - 2e^(-x) - 3x$ ?

Question

6801 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-29T02:26:13

Correction: That is not a logarithmic function, but here is how to find the critical numbers for the function.

$f(x) = e^x - 2e^(-x) - 3x$

Note that $"Dom:(f) = (-oo,oo)$

$f'(x) = e^x-2e^(-x)(-1)-3$

$= e^x+2e^(-x)-3$

$f'(x)$ is never undefined, so we need only find its zeros.

We need to solve

$e^x+2e^(-x)-3 = 0$

As a first step, try getting rid of negative exponents:

$e^x+2/e^x-3 = 0$

Get a common denomiator, noting that $e^x*e^x = e^(2x)$ , so we have

$(e^(2x)+2-3e^x)/e^x = 0$ .

So now we need to solve:

$e^(2x)-3e^x+2 = 0$

Again note that $e^(2x)=(e^x)^2$ , so we can factor:

$(e^x-1)(e^x-2) = 0$

$e^x=1$ $" "$ OR $" "$ $e^x=2$

$x=0$ $" "$ OR $" "$ $x=ln2$

Both are in $"Dom"(f)$ , so both are critical numbers.

How do you find critical points of a logarithmic function y = e^x - 2e^(-x) - 3xy = e^x - 2e^(-x) - 3x?