5.6 Minimum and maximum values  (Page 6/6)

Problem : Find minimum value of modulus function

Solution : Modulus function is defined as :

| x ; x≥0 f(x) = || -x ; x<0

For x>0,

$$f\left(x\right)=x$$ $$\Rightarrow f\prime \left(x\right)=1>0$$

Since first derivative is positive, given function is increasing function for x>0.

For x<0,

$$f\left(x\right)=-x$$ $$\Rightarrow f\prime \left(x\right)=-1<0$$

Since first derivative is negative, given function is decreasing function for x<0. Overall sign diagram of modulus function is as shown here :

At x=0, the function is decreasing to its left and increasing to its right. It means function has minimum at x=0.

$$\text{Minimum value}=0$$

Note that minimum value in this case is also least value as there is only one minimum in the entire domain. Hence, minimum at x=0 is global minimum.

Problem : The function $y=a{\log }_{e}x+b{x}^2+x$ has exteme values at x=-1 and x=2. Find values of “a” and “b”.

Solution : Here extreme values (maximum or minimum) are given. We know that first derivative is zero at extreme values. Now,

$$\Rightarrow y\prime =aX\frac{1}{x}+2bx+1$$

At x =-1,

$$\Rightarrow y\prime =aX\frac{1}{-1}+2bX-1+1=0$$ $$\Rightarrow -a-2b+1=0$$

At x = 2,

$$\Rightarrow y\prime =aX\frac{1}{2}+2bX2+1=0$$ $$\Rightarrow \frac{a}{2}+4b+1=0$$ $$\Rightarrow a+8b+2=0$$

Solving two simultaneous equations,

$$\Rightarrow a=2$$ $$\Rightarrow b=-\frac{1}{2}$$

Problem : Find maximum and minimum values of function :

$$y=\frac{x^2-7x+6}{x-10}$$

Solution : This function is not defined for x=10. The function is continuous except at this point. Thus, minimum and maximum obtained do not belong to a continuous domain.

$$\Rightarrow y\prime =\frac{\left(x-10\right)\left(2x-7\right)-\left(x^2-7x+6\right)}{{\left(x-10\right)}^2}=\frac{2x^2-27x+70-x^2+7x-6}{{\left(x-10\right)}^2}$$

$$\Rightarrow y\prime =\frac{x^2-20x+64}{{\left(x-10\right)}^2}$$

Now denominator is a positive number for all x. Thus, sign diagram of first derivative is same as that of numerator. In order to draw sign diagram, we need to factorize numerator.

$$\Rightarrow x^2-20x+64=\left(x-4\right)\left(x-16\right)$$

Hence, critical points are 4 and 16. The sign diagram is as shown in the figure. At x=4, the function is increasing to its left and decreasing to its right. It means function has maximum at x=4.

$$\text{Maximum value}=\frac{x^2-7x+6}{x-10}=\frac{4^2-7X4+6}{4-10}=1$$

At x=16, the function is decreasing to its left and increasing to its right. It means function has minimum at x=16.

$$\text{Minimum value}=\frac{x^2-7x+6}{x-10}=\frac{{16}^2-7X16+6}{16-10}=25$$

Note that minimum value is greater than maximum value.