How do you find the exact minimum value of #f(x) = e^x + e^(-2x)# on [0,1]?

Question

Ezra Adams · Accepted Answer

See the explanation

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]}

Samantha Adkison · Answer

undefined To find the exact minimum value of $ f(x) = e^x + e^{-2x} $ on the interval $[0,1]$, you need to first find the critical points by taking the derivative of $ f(x) $ and setting it equal to zero. Then, you evaluate $ f(x) $ at these critical points and the endpoints of the interval $[0,1]$, and the smallest value among these will be the minimum value of $ f(x) $ on the interval.

1. Take the derivative of $ f(x) $ with respect to $ x $:
$$ f'(x) = e^x - 2e^{-2x} $$

2. Set $ f'(x) $ equal to zero and solve for $ x $ to find the critical points:
$$ e^x - 2e^{-2x} = 0 $$
$$ e^x = 2e^{-2x} $$
$$ e^x = 2e^{-(2x)} $$
$$ e^x = 2e^x e^{-2x} $$
$$ e^x = 2e^{-x} $$
$$ e^{2x} = 2 $$
$$ 2x = \ln(2) $$
$$ x = \frac{1}{2} \ln(2) $$

3. Evaluate $ f(x) $ at the critical point $ x = \frac{1}{2} \ln(2) $ and the endpoints of the interval $[0,1]$:
$$ f(0) = e^0 + e^{-2*0} = 1 + 1 = 2 $$
$$ f(1) = e^1 + e^{-2*1} = e + e^{-2} $$
$$ f\left(\frac{1}{2} \ln(2)\right) = e^{\frac{1}{2} \ln(2)} + e^{-2 \cdot \frac{1}{2} \ln(2)} = 2^{\frac{1}{2}} + 2^{-1} = \sqrt{2} + \frac{1}{2} $$

4. Compare the values obtained:
$$ f(0) = 2 $$
$$ f\left(\frac{1}{2} \ln(2)\right) = \sqrt{2} + \frac{1}{2} $$
$$ f(1) = e + e^{-2} $$

The minimum value of $ f(x) $ on the interval $[0,1]$ is $ 2 $.