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Question: The absolute maximum of \[{{x}^{40}}-{{x}^{20}}\] on the interval \[[0,1]\] is 1) \[\dfrac{-1}{4}\...

The absolute maximum of x40−x20{{x}^{40}}-{{x}^{20}} on the interval [0,1][0,1] is

  1. −14\dfrac{-1}{4}
  2. 00
  3. 14\dfrac{1}{4}
  4. 12\dfrac{1}{2}
Explanation

Solution

In this type of question you need to find the first derivative equation of the function and then put that derivative equals to zero, this equation will give you points of local maxima, among all those the highest one and it must satisfy the given domain interval.

Complete step by step answer:
First we are going to differentiate one time the given function, because the first derivative of a function shows the nature of the slope of that point. So when this is equal to zero there is no slope, because the slope of the minimum point on the graph is equal to zero, and as you must be knowing that the derivative of a function is just the slope of the curve of the function.
Let say the function be f(x)=x40−x20f(x)={{x}^{40}}-{{x}^{20}}
So, on differentiating with respect to xx :
Here, to differentiate we will use the formula:
xn=nxn−1{{x}^{n}}=n{{x}^{n-1}}
f′(x)=40x39−20x19f'(x)=40{{x}^{39}}-20{{x}^{19}}
Now, we have got the required derivative, now we will put this derivative equal to zero.
Because the derivative provides information about the gradient or slope of the graph of a function we can use it to locate points on a graph where the gradient is zero to find maxima and minima.
Therefore,
⇒f′(x)=0\Rightarrow f'(x)=0
⇒40x39−20x19=0\Rightarrow 40{{x}^{39}}-20{{x}^{19}}=0
On taking common 20x1920{{x}^{19}}from above equation we got,
⇒20x19(2x20−1)=0\Rightarrow 20{{x}^{19}}(2{{x}^{20}}-1)=0
Since we have the equation like a.b=0a.b=0
So, possibilities are a=0a=0 or b=0b=0
Therefore, on taking one factor equal to zero we got,

& \Rightarrow 20{{x}^{19}}=0 \\\ & \Rightarrow x=0 \\\ \end{aligned}$$ And on taking other factor equals to zero we got, $$\begin{aligned} & \Rightarrow 2{{x}^{20}}-1=0 \\\ & \Rightarrow {{x}^{20}}=\dfrac{1}{2} \\\ \end{aligned}$$ Here, although we can find further value of $$x$$ by solving the above equation further, we are leaving it as it is to make our calculation simple. So, we can find now the values of function $$f(x)$$ for above value of $$x$$ we found at which function will give the different local maxima, When $$x=0$$ , the value of function is: $$f(x)={{0}^{40}}-{{0}^{20}}$$ $$\Rightarrow f(x)=0$$ When $$x=\dfrac{1}{{{2}^{\dfrac{1}{20}}}}$$ i.e. when $${{x}^{20}}=\dfrac{1}{2}$$ , the value of function is: $$f(x)={{x}^{20}}({{x}^{20}}-1)$$ $$\Rightarrow f(x)=\dfrac{1}{2}\left( \dfrac{1}{2}-1 \right)$$ $$\Rightarrow f(x)=\dfrac{-1}{4}$$ So, the absolute maximum of the given function is at $$f(x)=0$$ . **So, the correct answer is “Option B”.** **Note:** Such applications of maxima and minima exist in economics, business, and engineering. Many can be solved using the methods of differential calculus described above. For example, in any manufacturing business it is usually possible to express profit as a function of the number of units sold.