Question

Number of turning point of y=(x+4)(x+2)(x-1)(x-3)

Answer 1

$$y=(x+4)(x+2)(x-1)(x-3)$$

$$y=(x^2+6x+8)(x^2-4x+3)$$

$$y'=(2x+6)(x^2-4x+3)+(2x-4)(x^2+6x+8)$$

$$=2x^3-8x^2+6x+6x^2-24x+18$$

$$+2x^3+12x^2+16x-4x^2-24x-32$$

$$=4x^3+6x^2-26x-14$$

Find the critical number(s)

$$y'=0=>4x^3+6x^2-26x-14=0$$

$$2x^3+3x^2-13x-7=0$$

$$(2x^3+x^2)+(2x^2+x)-(14x+7)=0$$

$$x^2(2x+1)+x(2x+1)-7(2x+1)=0$$

$$(2x+1)(x^2+x-7)=0$$

$$2x+1=0=>x_1=-\dfrac{1}{2}$$

$$x^2+x-7=0$$

$$D=(1)^2-4(1)(-7)=29>0$$

$$x_2=\dfrac{-1-\sqrt{29}}{2}, x_3=\dfrac{-1+\sqrt{29}}{2}$$

Critical numbers:

$$-\dfrac{1}{2}, \dfrac{-1-\sqrt{29}}{2},\dfrac{-1+\sqrt{29}}{2}$$

If $x<\dfrac{-1-\sqrt{29}}{2}, y'<0, y$ decreases.

If $\dfrac{-1-\sqrt{29}}{2}<x<-\dfrac{1}{2}, y'>0, y$ increases.

If $-\dfrac{1}{2}<x<\dfrac{-1+\sqrt{29}}{2}, y'<0, y$ decreases.

If $x>\dfrac{-1+\sqrt{29}}{2}, y'>0, y$ increases.

Then the function $y=(x+4)(x+2)(x-1)(x-3)$ has $4-1=3$ turning points.