Chapter 4.1, Problem 53E

(a)

To determine

To show : that $f$ can have 0, 1 or 2 critical points.

Explanation of Solution

Given information :

  $f\left(x\right)=a{x}^3+b{x}^2+cx+d$

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

Differentiate $f\left(x\right)=a{x}^3+b{x}^2+cx+d$ with respect to $x$ .

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(a{x}^3+b{x}^2+cx+d\right)\\ =3a{x}^2+2bx+c\end{array}$

For critical points, $f^{\prime }\left(x\right)=0$ then $3a{x}^2+2bx+c=0$ .

If $a=0$ and $b=0$ , then $c=0$ which is not possible. No critical point exists.

If ${\left(2b\right)}^2-4\left(3a\right)c=0$ , then only one critical point exists.

If ${\left(2b\right)}^2-4\left(3a\right)c>0$ , then two critical point exists.

No critical point,

  

One critical point,

  

Two critical point,

  

(b)

To determine

To find : local extreme values of $f$ .

Answer to Problem 53E

No critical point, no extreme value.

one critical point, one extreme value.

two critical point exists, two extreme values.

Explanation of Solution

Given information :

  $f\left(x\right)=a{x}^3+b{x}^2+cx+d$

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

Differentiate $f\left(x\right)=a{x}^3+b{x}^2+cx+d$ with respect to $x$ .

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(a{x}^3+b{x}^2+cx+d\right)\\ =3a{x}^2+2bx+c\end{array}$

For critical points, $f^{\prime }\left(x\right)=0$ then $3a{x}^2+2bx+c=0$ .

If $a=0$ and $b=0$ , then $c=0$ which is not possible. No critical point, no extreme value.

If ${\left(2b\right)}^2-4\left(3a\right)c=0$ , then only one critical point, one extreme value.

If ${\left(2b\right)}^2-4\left(3a\right)c>0$ , then two critical point exists, two extreme values.

End points not included.