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

Question

6443 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.