Question

Find the maximum and minimum value of the following expression ${{\left( x-2 \right)}^{6}}{{\left( x-3 \right)}^{5}}$

Answer 1

Hint: Now we know that for a function f(x) the conditions for extrema is $f'\left( x \right)=0$. Now this extrema can be a minimum or a maximum. Now we will substitute the value of x in f(x). The x which gives greater value is maximum the x which gives lesser value is minimum.

Complete step by step answer:
Now consider the given function $f\left( x \right)={{\left( x-2 \right)}^{6}}{{\left( x-3 \right)}^{5}}$.
Now we know that $\left( f\left( x \right).g\left( x \right) \right)'=f'\left( x \right)g\left( x \right)+f\left( x \right)g'\left( x \right)$
Hence with this formula if we differentiate the above expression we get.
$f'\left( x \right)=\dfrac{d{{\left( x-2 \right)}^{6}}}{dx}{{\left( x-3 \right)}^{5}}+{{\left( x-2 \right)}^{6}}\dfrac{d{{\left( x-3 \right)}^{5}}}{dx}.........................(1)$
Now we know that $\left( f\left( g\left( x \right) \right) \right)'=f'\left( g\left( x \right) \right).g'\left( x \right)$
Hence we have
$\dfrac{d{{\left( x-2 \right)}^{6}}}{dx}=6{{\left( x-2 \right)}^{5}}\left( 1 \right)=6{{\left( x-2 \right)}^{5}}$ and $\dfrac{d{{\left( x-3 \right)}^{5}}}{dx}=5{{\left( x-3 \right)}^{4}}\left( 1 \right)=5{{\left( x-3 \right)}^{4}}$
Now substituting this in equation (1) we get.
$f'\left( x \right)=6{{\left( x-2 \right)}^{5}}{{\left( x-3 \right)}^{5}}+5{{\left( x-2 \right)}^{6}}{{\left( x-3 \right)}^{4}}$
Now we will equate this to 0 we get the conditions for extremum.
$6{{\left( x-2 \right)}^{5}}{{\left( x-3 \right)}^{5}}+5{{\left( x-2 \right)}^{6}}{{\left( x-3 \right)}^{4}}=0$
Now taking ${{\left( x-2 \right)}^{5}}{{\left( x-3 \right)}^{4}}$ common from each term we get
$\begin{align}
  & \Rightarrow {{\left( x-2 \right)}^{5}}{{\left( x-3 \right)}^{4}}\left[ 6\left( x-3 \right)+5\left( x-2 \right) \right]=0 \\
 & \Rightarrow {{\left( x-2 \right)}^{5}}{{\left( x-3 \right)}^{4}}[6x-18+5x-10] \\
 & \Rightarrow {{\left( x-2 \right)}^{5}}{{\left( x-3 \right)}^{4}}\left[ 11x-28 \right]=0 \\
\end{align}$
Hence we get $f'\left( x \right)=0$ for x = 3, x = 2, and $x=\dfrac{28}{11}$ .
Hence we have the points of extremum are x = 3, x = 2, and $x=\dfrac{28}{11}$ .
Now let us check the value of function at each points of extrema
We have $f\left( x \right)={{\left( x-2 \right)}^{6}}{{\left( x-3 \right)}^{5}}$
Hence $f\left( 2 \right)=f\left( 3 \right)=0............(2)$ .
Now we have $\dfrac{28}{11}=2.54$
Now we know that 2.54 < 3
Hence we get $\left( 2.54-3 \right) < 0$
Now the odd power of a negative number is negative.
Hence ${{\left( 2.54-3 \right)}^{5}}<0...............\left( 3 \right)$
But we have ${{\left( 2.54-2 \right)}^{6}}>0...............\left( 4 \right)$
Now we know that $\left( + \right)\times \left( - \right)=\left( - \right)$ hence we get
$\begin{align}
  & {{\left( 2.54-3 \right)}^{5}}{{\left( 2.54-2 \right)}^{6}}<0 \\
 & \Rightarrow f\left( 2.54 \right)<0 \\
 & \Rightarrow f\left( \dfrac{28}{11} \right)<0 \\
\end{align}$
Now from equation (2) we have that
\[f\left( \dfrac{28}{11} \right) < f\left( 2 \right)\] and $f\left( \dfrac{28}{11} \right) < f\left( 3 \right)$

Hence we have maximum at x = 2 and x = 3 and minimum at x = $\dfrac{28}{11}$

Note: We can also check if the extremum is maximum or minimum by second derivative test. Let us say if x is a point of extremum then if at this x we have $f''\left( x \right)>0$ then it is a minimum and if $f''\left( x \right)<0$ then it is a maximum. Hence we can find the maximum and minimum of the given expression.