Question

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

Answer 1

Hint: 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.