Question

How do you find the exact minimum value of $f\left( x \right) = {e^x} + {e^{ - 2x}}$ on $\left[ {0,1} \right]$?

Answer 1

Hint: First, we have to find the differentiation of $f$ with respect to $x$ using differentiation rules. Next, find all critical points of $f$ in the interval, i.e., find points $x$ where either $f'\left( x \right) = 0$ or $f$ is not differentiable using exponential and logarithm properties. Next, evaluate the value of $f$ at critical points and at the end points of the interval $\left[ {0,1} \right]$. Next, identify the minimum value of $f$ out of the values calculated. The minimum value will be the absolute minimum (least) value of $f$.
If $f\left( x \right)$and $g\left( x \right)$are differentiable functions and c is a constant.
1. $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$
2. $\dfrac{{d\left( c \right)}}{{dx}} = 0$
3. $\dfrac{d}{{dx}}\left\{ {c \cdot f\left( x \right)} \right\} = c \times \dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$
4. $\dfrac{d}{{dx}}\left[ {f\left( x \right) \cdot g\left( x \right)} \right] = f\left( x \right)\dfrac{d}{{dx}}g\left( x \right) + g\left( x \right)\dfrac{d}{{dx}}f\left( x \right)$
5. $\dfrac{d}{{dx}}\left[ {f\left( x \right) \pm g\left( x \right)} \right] = \dfrac{d}{{dx}}f\left( x \right) \pm \dfrac{d}{{dx}}g\left( x \right)$
6. $\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$

Complete step by step answer:
Given function is $f\left( x \right) = {e^x} + {e^{ - 2x}}$.
We have to find the exact minimum value of $f\left( x \right) = {e^x} + {e^{ - 2x}}$ on $\left[ {0,1} \right]$.
First, we have to find the differentiation of $f$ with respect to $x$.
So, differentiating $f$ with respect to $x$, we get
$\dfrac{d}{{dx}}\left( {f\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {{e^x} + {e^{ - 2x}}} \right)$
Use the property $\dfrac{d}{{dx}}\left[ {f\left( x \right) \pm g\left( x \right)} \right] = \dfrac{d}{{dx}}f\left( x \right) \pm \dfrac{d}{{dx}}g\left( x \right)$ in above equation where $f\left( x \right)$and $g\left( x \right)$are differentiable functions.
$ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {{e^x}} \right) + \dfrac{d}{{dx}}\left( {{e^{ - 2x}}} \right)$
Use the property $\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$ in above equation.
$ \Rightarrow f'\left( x \right) = {e^x} + {e^{ - 2x}}\dfrac{d}{{dx}}\left( { - 2x} \right)$
Use the property $\dfrac{d}{{dx}}\left\{ {c \cdot f\left( x \right)} \right\} = c \times \dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$ in above equation where $f\left( x \right)$is a differentiable function and c is a constant.
$ \Rightarrow f'\left( x \right) = {e^x} - 2{e^{ - 2x}}\dfrac{d}{{dx}}\left( x \right)$
Use the property $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$ in above equation where $f\left( x \right)$is a differentiable function.
$ \Rightarrow f'\left( x \right) = {e^x} - 2{e^{ - 2x}}$
Now, find all critical points of $f$ in the interval, i.e., find points $x$ where either $f'\left( x \right) = 0$ or $f$ is not differentiable.
So, put $f'\left( x \right) = 0$, we get
${e^x} - 2{e^{ - 2x}} = 0$
${e^x} = \dfrac{2}{{{e^{2x}}}}$
Now, use the property ${e^{a + b}} = {e^a}{e^b}$ in the above equation.
${e^{3x}} = 2$
Take logarithm on both sides of the equation.
$\ln \left( {{e^{3x}}} \right) = \ln \left( 2 \right)$
Use property $\ln \left( {{m^n}} \right) = n\ln \left( m \right)$ and $\ln \left( e \right) = 1$ in above equation.
$3x\ln \left( e \right) = \ln \left( 2 \right)$
$ \Rightarrow x = \dfrac{{\ln \left( 2 \right)}}{3}$
or $x = \ln \sqrt[3]{2}$
Thus, $x = \ln \sqrt[3]{2}$ is a critical point of $f$.
Now, we have to evaluate the value of $f$ at critical point and at the end points of the interval $\left[ {0,1} \right]$,
 i.e., at $x = 0$, $x = \ln \sqrt[3]{2}$ and $x = 1$
$f\left( 0 \right) = {e^0} + {e^{ - 2 \times 0}} = 1 + 1 = 2$
$f\left( {\ln \sqrt[3]{2}} \right) = {e^{\ln \sqrt[3]{2}}} + {e^{ - 2 \times \ln \sqrt[3]{2}}} = \sqrt[3]{2} + {e^{\ln \dfrac{1}{{{{\left( {\sqrt[3]{2}} \right)}^2}}}}} = \sqrt[3]{2} + \dfrac{1}{{{{\left( {\sqrt[3]{2}} \right)}^2}}} = \sqrt[3]{2} + \dfrac{1}{{\sqrt[3]{4}}} = 1.889881575$
$f\left( 1 \right) = {e^1} + {e^{ - 2 \times 1}} = e + \dfrac{1}{{{e^2}}} = 2.853617112$
Now, identify the maximum and minimum values of $f$ out of the values calculated. The maximum value will be the absolute maximum (greatest) value of $f$ and the minimum value will be the absolute minimum (least) value of $f$.
The absolute maximum value of $f$ on $\left[ {0,1} \right]$ is $2.853617112$, occurring at $x = 1$, and absolute minimum value of $f$ on $\left[ {0,1} \right]$ is $1.889881575$ which occurs at $x = \ln \sqrt[3]{2}$.

Therefore, the exact minimum value of $f\left( x \right) = {e^x} + {e^{ - 2x}}$ on $\left[ {0,1} \right]$ is $\sqrt[3]{2} + \dfrac{1}{{\sqrt[3]{4}}}$ or $1.889881575$.

Note: Algorithm for finding the maximum and the minimum values of a function in a closed interval.
Step 1: Find all critical points $f$ in the interval, i.e., find points $x$ where either $f'\left( x \right) = 0$ or $f$ is not differentiable.
Step 2: Take the end points of the interval.
Step 3: At all these points (listed in Step 1 and 2), calculate the values of $f$.
Step 4: Identify the maximum and minimum values of $f$ out of the values calculated in Step 3. The maximum value will be the absolute maximum (greatest) value of $f$ and the minimum value will be the absolute minimum (least) value of $f$.