Chapter 5.1, Problem 53E

a.

To determine

To show: that function have $0,1\text{ or 2}$ critical points.

Explanation of Solution

Given information:

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

Proof: to find the extreme value, differentiate the function because function has no end points,

  $\begin{array}{l}f\left(x\right)=a{x}^3+b{x}^2+cx+d\\ f\text{'}\left(x\right)=3a{x}^2+2bx+c\text{ [differentiate and simplify] }\end{array}$

Now, equal the differentiation to zero,

  $3a{x}^2+2bx+c=0$

Form the above equation, if $c=0$ then $0$ is a critical points of the function and if $3a+2b+c=0$ then $1$ is a critical points of the function and if $12a+4b+c=0$ then $2$ is a critical points of the function.

Example: the function $f\left(x\right)=x^3-3x$ has two critical points at $x=-1\text{ and }x=1$

  

The function $f\left(x\right)=x^3-1$ has one critical point at $x=0$ .

  

The function $f\left(x\right)=x^3+x$ has no critical points.

b.

To determine

To find: that how many local extreme function can have.

Answer to Problem 53E

The function can have two or none local extreme.

Explanation of Solution

Given information:

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

Calculation: to find the extreme value, differentiate the function because function has no end points,

  $\begin{array}{l}f\left(x\right)=a{x}^3+b{x}^2+cx+d\\ f\text{'}\left(x\right)=3a{x}^2+2bx+c\text{ [differentiate and simplify] }\end{array}$

Now, equal the differentiation to zero,

  $3a{x}^2+2bx+c=0$

Because, the differentiation of the function is an quadratic equation, so the value of the local extreme can be two or none.