Chapter 2.2, Problem 45E

(a)

To determine

To find: all the real zeros of the given function.

Answer to Problem 45E

The real zeros are $t=0,\pm \sqrt{3}$

Explanation of Solution

Given information:

Consider $g(t)=t^5-6t^3+9t$

Calculation:

Find the zero’s of the given function

The function

  $\begin{array}{l}g(t)=t^5-6t^3+9t=0\\ t(t^4-6t^2+9)=0\\ t=0\\ t^4-6t^2+9=0\\ $ \Rightarrow {(t^2-3)}^2=0$t^2=3\\ t=\pm \sqrt{3}\end{array}$

Hence, the real zeros are $t=0,\pm \sqrt{3}$

(b)

To determine

To find: whether the multiplicity of the each zero of the given function is even or odd .

Answer to Problem 45E

Multiplicity of $t=0$ is odd and

Multiplicity of $t=0,\pm \sqrt{3}$ is even.

Explanation of Solution

Given information:

Consider $g(t)=t^5-6t^3+9t$

Calculation:

From part a,

The real zeros are $t=0,\pm \sqrt{3}$

Multiplicity of $t=0$ is $=1$

Multiplicity of $t=\pm \sqrt{3}$ is $=2$

Hence, Multiplicity of $t=0$ is odd and

Multiplicity of $t=0,\pm \sqrt{3}$ is even

(c)

To determine

To find: the maximum possible number of turning points of the given function.

Answer to Problem 45E

The maximum number of turning points $=4$

Explanation of Solution

Given information:

Consider $g(t)=t^5-6t^3+9t$

Calculation:

The maximum number of turning points is one less than the degree of the function

Degree of the function is $=5$

Hence, the maximum number of turning points $=4$

(d)

To determine

To plot: the given function and to verify the results of previous parts.

Explanation of Solution

Given information:

Consider $g(t)=t^5-6t^3+9t$

Graph:

The graph of the $g(t)=t^5-6t^3+9t$ is shown below

  

Hence, the results of previous parts are correct from the graph.