### Application Of Derivatives Question 13

#### Question 13 - 01 February - Shift 2

The sum of the abosolute maximum and minimum values of the function $f(x)=|x^{2}-5 x+6|-3 x+2$ in the interval $[-1,3]$ is equal to :

(1) 10

(2) 12

(3) 13

(4) 24

## Show Answer

#### Answer: (1)

#### Solution:

#### Formula: Maximum of function and minima of a function

$ \begin{aligned} & f(x)=|x^{2}-5 x+6|-3 x+2 \\ & f(x)= \begin{cases}x^{2}-8 x+8 & ; x \in[-1,2] \\ -x^{2}+2 x-4 & ; x \in[2,3]\end{cases} \end{aligned} $

$ \begin{aligned} & f^{\prime}(x)=2 x-8 \\ & f^{\prime}(x)=0 \\ & 2 x-8=0 \\ & 2 x=8 \\ & x=4 \end{aligned} $

$\Rightarrow$ Since $x=4$ is outside the interval $[-1,3]$, there are no critical points within the interval.

Evaluate $f(x)$ at the endpoints of the interval: For $x=-1$

$ f(-1)=(-1)^2-5(-1)+6-3(-1)+2=1+5+6+3+2=17 $

For $x=3$

$ f(3)=3^2-5(3)+6-3(3)+2=9-15+6-9+2=-7 $

The absolute maximum value is 17 (at $x=-1$ ), and the absolute minimum value is -7 (at $x=3$ ).

The sum of the absolute maximum and minimum values is: $17+(-7)=10$

So, the correct option is (1)