Question

The minimum value of $4{{e}^{2x}}+9{{e}^{-2x}}$ is
1. $11$
2. $12$
3. $10$
4. $14$

Answer 1

At first we have to differentiate the function and then equating it to 0 we get to know the points or values at 0 where the function can be minimum or maxima. Then after finding those values we find the second derivative and check values for which the first derivative is 0. If the value for x in the second derivative is positive the value refers to minimum and if the value for x is negative then it refers to the maximum of the function.

Complete step-by-step solution:
First we will start by knowing and finding out what the expression is therefore
$f(x)=4{{e}^{2x}}+9{{e}^{-2x}}$
To solve further we must find out the first derivative of f(x) therefore we can solve and find that it will be
$f'(x)=8{{e}^{2x}}-18{{e}^{-2x}}$
Now to find the values at which we can compare and see if it can be positive or negative we must equate the first derivative to zero to get the answer therefore
$8{{e}^{2x}}-18{{e}^{-2x}}=0$
Simplifying we get
${{e}^{2x}}=\dfrac{3}{2}$
Taking log on both sides to find the value of x
$x=\log {{\left( \dfrac{3}{2} \right)}^{\dfrac{1}{2}}}$
Now to find if the function is at minima or at maxima we take the second derivative
$f''(x)=16{{e}^{2x}}+36{{e}^{-2x}}$
Putting the value for x
$f''\left( \log {{\left( \dfrac{3}{2} \right)}^{\dfrac{1}{2}}} \right)=\dfrac{4\times 3}{2}+\dfrac{9\times 2}{3}$
Simplifying
$=6+6$
$=12$
Since it is positive i.e. greater than zero therefore it is the minimum value of the given function.

Note: If the second derivative is found to be zero it means that it is a point of inflection that is at the point the second derivative will change its sign. Now we can say that minima is the point where any function reaches its lowest point and maxima is the point on the function where it reaches the highest point. If a function is in relation to x the minima and maxima will lie on the y axis and vice versa.