Question

How do you find the maximum or minimum of $f(x)={{x}^{2}}-8x+2$?

Answer 1

This type of question is based on the concept of maxima and minima. We have to first consider the given function and differentiate f(x) with respect to x. Equate ${f}'(x)$ to 0 and find the value of x, where ${f}'(x)$ is the derivative of f(x). Here, x=4. Now differentiate ${f}'(x)$with respect to x. if the value of ${f}''(x)$ is less than 0, then the function have local maximum at x=4 and if ${f}''(x)$ is more than 0, then the function have local minimum at x=4. Here, ${f}''(x)>0$ and thus the function has a local minimum. To find the minimum of $f(x)={{x}^{2}}-8x+2$, we have to substitute x=4 in the function which is the required answer.

Complete step-by-step solution:
According to the question, we are asked to find the maximum or minimum of $f(x)={{x}^{2}}-8x+2$.
We have been given the function $f(x)={{x}^{2}}-8x+2$. ----------(1)
First, differentiate f(x) with respect to x.
$\dfrac{d}{dx}\left( f(x) \right)=\dfrac{d}{dx}\left( {{x}^{2}}-8x+2 \right)$
We have to use addition rule of differentiation, that is, $\dfrac{d}{dx}\left( u+v+w \right)=\dfrac{du}{dx}+\dfrac{dv}{dx}+\dfrac{dw}{dx}$.
$\Rightarrow \dfrac{d}{dx}\left( f(x) \right)=\dfrac{d}{dx}\left( {{x}^{2}} \right)+\dfrac{d}{dx}\left( -8x \right)+\dfrac{d}{dx}\left( 2 \right)$
Since the differentiation of a constant is zero, $\dfrac{d}{dx}\left( 2 \right)=0$.
$\Rightarrow \dfrac{d}{dx}\left( f(x) \right)=\dfrac{d}{dx}\left( {{x}^{2}} \right)+\dfrac{d}{dx}\left( -8x \right)$
On taking the constants out from $\dfrac{d}{dx}\left( -8x \right)$, we get
$\Rightarrow \dfrac{d}{dx}\left( f(x) \right)=\dfrac{d}{dx}\left( {{x}^{2}} \right)-8\dfrac{dx}{dx}$
We know that $\dfrac{dx}{dx}=1$, we get
$\Rightarrow \dfrac{d}{dx}\left( f(x) \right)=\dfrac{d}{dx}\left( {{x}^{2}} \right)-8$
Using the power rule of differentiation, that is, $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$, we get
$\dfrac{d}{dx}\left( f(x) \right)=2{{x}^{2-1}}-8$
$\Rightarrow \dfrac{d}{dx}\left( f(x) \right)=2x-8$
$\therefore {f}'\left( x \right)=2x-8$
Now, equate ${f}'\left( x \right)$ to 0.
$\Rightarrow 2x-8=0$
Add 8 on both the sides of the equation. We get
$2x-8+8=0+8$
On further simplification, we get
$\Rightarrow 2x=8$
Divide the whole equation by 2.
$\Rightarrow \dfrac{2x}{2}=\dfrac{8}{2}$
Cancelling out the common terms, we get
$x=4$
Now, we have to find the second derivative of f(x), that is, $\dfrac{d}{dx}\left( {f}'\left( x \right) \right)$.
Therefore, $\dfrac{d}{dx}\left( {f}'\left( x \right) \right)=\dfrac{d}{dx}\left( 2x+8 \right)$
We have to use addition rule of differentiation, that is, $\dfrac{d}{dx}\left( u+v \right)=\dfrac{du}{dx}+\dfrac{dv}{dx}$.
$\Rightarrow \dfrac{d}{dx}\left( {f}'\left( x \right) \right)=\dfrac{d}{dx}\left( 2x \right)+\dfrac{d}{dx}\left( 8 \right)$
Since differentiation of a constant is 0, we get
$\dfrac{d}{dx}\left( {f}'\left( x \right) \right)=\dfrac{d}{dx}\left( 2x \right)$
Taking the constants out of the differentiation, we get
$\Rightarrow \dfrac{d}{dx}\left( {f}'\left( x \right) \right)=2\dfrac{dx}{dx}$
We know that $\dfrac{dx}{dx}=1$, we get
$\dfrac{d}{dx}\left( {f}'\left( x \right) \right)=2$
$\therefore {f}''\left( x \right)=2$
Here, ${f}''\left( x \right)>0$. Since ${f}''\left( x \right)>0$, the function is minimum at x=4.
We now found that the function has minimum value.
To find the minimum value of f(x), we have to substitute x=4 in f(x).
$f(x=4)={{4}^{2}}-8\times 4+2$
On further simplification, we get
$f(x=4)=16-8\times 4+2$
$\Rightarrow f(x=4)=16-32+2$
$\Rightarrow f(x=4)=16-30$
$\Rightarrow f(x=4)=-14$
Therefore, the value of f(x) at x=4 is -14.
Hence, the minimum value of $f(x)={{x}^{2}}-8x+2$ is -14.

Note: We should not get confused in the second derivative of f(x). If ${f}''(x)>0$, then the function has local minimum value and if ${f}''(x)<0$, then the function has local maximum value and not vice-versa. Differentiation of a constant is not x but is 0. Avoid calculation mistakes based on sign conventions.