5.6 Minimum and maximum values  (Page 5/6)

Problem : Determine minimum and maximum values of function :

$$f\left(x\right)=\frac{x^3}{3}-x$$

Solution : Differentiating with respect x, we have :

$$\Rightarrow f\prime \left(x\right)=\frac{1}{3}X3x^2-1=x^2-1=\left(x-1\right)\left(x+1\right)$$

The roots of the corresponding equation are -1 and 1. Now, differentiating with respect to x again,

$$\Rightarrow f\prime \prime \left(x\right)=2x$$

Putting, $x=-\mathrm{1,}f\prime \prime \left(x\right)=-2<0$ . Hence, function has maximum value at x=-1.

$$\text{Maximum value}=\frac{{\left(-1\right)}^3}{3}-\left(-1\right)=-\frac{1}{3}+1=\frac{2}{3}$$

Putting, $x=\mathrm{1,}f\prime \prime \left(x\right)=2>0$ . Hence, function has minimum value at x=1.

$$\text{Minimum value}=\frac{{\left(1\right)}^3}{3}-1=\frac{1}{3}-1=-\frac{2}{3}$$

Problem : Determine minimum and maximum values of function :

$$2x^3-9x^2+12x-11$$

Solution : Differentiating with respect x, we have :

$$\Rightarrow f\prime \left(x\right)=6x^2-18x+12=6\left(x^2-3x+2\right)=6\left(x-1\right)\left(x-2\right)$$

The roots are 1 and 2. Now, differentiating with respect to x again,

$$\Rightarrow f\prime \prime \left(x\right)=12x-18$$

Putting, $x=\mathrm{1,}f\prime \prime \left(x\right)=-6<0$ . Hence, function has maximum value at x=1.

$$\text{Maximum value}=2x^3-9x^2+12x-10=2X{1}^3-9X{1}^2+12X1-11=-6$$

Putting, $x=\mathrm{2,}f\prime \prime \left(x\right)=6>0$ . Hence, function has minimum value at x=2.

$$\text{Minimum value}=2x^3-9x^2+12x-11=2X{2}^3-9X{2}^2+12X2-11=16-36+24-11=-7$$

Problem : Determine minimum and maximum values of function :

$$f\left(x\right)=x+\frac{1}{x}$$

Solution :

$$\Rightarrow f\prime \left(x\right)=1-\frac{1}{x^2}$$

$$\Rightarrow f\prime \left(x\right)=\frac{x^2-1}{x^2}$$

The roots are -1 and 1. Now, differentiating with respect to x again,

$$\Rightarrow f\prime \prime \left(x\right)=\frac{2}{x^3}$$

At $x=\mathrm{1,}f\prime \prime \left(x\right)=2>0$ . Function has minimum value of 2 at x=1. At $x=-\mathrm{1,}f\prime \prime \left(x\right)=-2<0$ . Function has maximum value of -2 at x=-1. Note that minimum value is greater than maximum value. It is possible as function is not defined at x=0. A maximum of smaller value exist left to it and a minimum of higher value exist to the right of it.

Problem : Determine minimum and maximum values of function :

$$f\left(x\right)=x^5-5x^4+5x^3-5$$

Solution : Differentiating with respect x, we have :

$$\Rightarrow f\prime \left(x\right)=5x^4-20x^3+15x^2$$

Equating to zero, we have :

$$5x^4-20x^3+15x^2=0$$ $$\Rightarrow x^4-4x^3+3x^2=0$$ $$\Rightarrow x^2\left(x^2-4x+3\right)=0$$ $$\Rightarrow x^2\left(x-1\right)\left(x-3\right)=0$$

The roots are 0, 1 and 3. Now, differentiating with respect to x again,

$$\Rightarrow f\prime \prime \left(x\right)=20x^3-60x^2+30x$$

Putting, $x=\mathrm{0,}f\prime \prime \left(x\right)=0$ . We need to differentiate again to evaluate this point. Putting, $x=\mathrm{1,}f\prime \prime \left(x\right)=-10<0$ . Hence, function has maximum value at x=1,

$$\text{Maximum value}=x^5-5x^4+5x^3-1=15-5X14+5X13-5=1-5+5-5=-4$$

Putting, $x=\mathrm{3,}f\prime \prime \left(x\right)=90>0$ . Hence, function has minimum value at x=3.

$$\text{Minimum value}=x^5-5x^4+5x^3-1=3^5-5X{3}^4+5X{3}^3-5=243-5X81+5X27-5=-32$$

In order to determine nature at point x=0, we differentiate again,

$$\Rightarrow f\prime \prime \prime \left(x\right)=60x^2-120x+30$$

Putting, $x=\mathrm{0,}f\prime \prime \prime \left(x\right)=30>0$ . Hence, function has neither minimum nor maximum value at x=0.