The minimum is either #f(0) = 2#
or #f(1) = e+e^-2 = e +1/e^2 =(e^3+1)/e^2#
or #f(c)# for some critical number of #f# in #[0,1]#.
So we need to find critical numbers for #f#
#f'(x) = e^x-2e^(-2x)#
# = e^x-2/e^(2x)#
# = (e^(3x)-2)/e^(2x)#
Since the denominator #e^(2x)# is never #0# (it is always positive), we see that #f'# is never undefined.
Solving #f'(x) = 0# gets us #e^(3x) = 2# so
#3x = ln2# and #x = ln2/3 = ln root(3)2#
Because #ln# is an increasing function, and #1 < root(3)2 < e#,
we see that #lnroot(3)2# is in #[0,1]#
So #c = lnroot(3)2 = ln2/3# is the only critical number.
We need to determine whether #f(c)# is a minimum, a maximum or neither.
The denominator of #f'(x)# is always positive, so the sign of #f'(x)# will be the same as the sign of the numerator:
#e^(3x)-2#.
For #x < ln2/3#, we have #3x < ln2#, and #e^(3x) < e^ln2 = 2#
so #e^(3x)-2#. and #f'(x)# is negative.
By similar reasoning, for #x > ln2/3#, we have #f'(x) >0#
Therefore #f(ln2/3)# is a local minimum.
Furthermore, #f(ln2/3) < f(0)# since #f# is decreasing on #[0, ln2/3)#
and #f(ln2/3) < f(1)# since #f# is increasing on #(ln2/3, 1]#
So on the interval #[0,1]# the value #f(ln2/3)# is the absolute minimum.
#f(lnroot(3)2) = e^(lnroot(3)2) + e^(-2(ln(2^(1/3)))#
# = e^(lnroot(3)2)+ e^ln(2^(-2/3))#
# = root(3)2 + 1/root(3)4#
The minimum value of #f(x) = e^x + e^(-2x)# on [0,1] is #root(3)2 + 1/root(3)4#
(Rewrite using algebra until you're happy with the way it looks. Personally, I like: #3/root(3)4#)
Now that we're finished, it might be nice to see the graph of #f#
graph{(y - e^x-e^(-2x))=0 [-0.965, 2.453, 1.486, 3.195]}
