Section 4.5 : The Shape of a Graph, Part I
6. For h(t)=50+40t3−5t4−4t5 answer each of the following questions.
- Identify the critical points of the function.
- Determine the intervals on which the function increases and decreases.
- Classify the critical points as relative maximums, relative minimums or neither.
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a Identify the critical points of the function. Show SolutionWe need the 1st derivative to get the critical points so here it is.
h′(t)=120t2−20t3−20t4=−20t2(t2+t−6)=−20t2(t+3)(t−2)Now, recall that critical points are where the derivative doesn’t exist or is zero. Clearly this derivative exists everywhere (it’s a polynomial….) and because we factored the derivative we can easily identify where the derivative is zero. The critical points of the function are,
\require{bbox} \bbox[2pt,border:1px solid black]{{t = - 3,\,\,\,\,\,\,\,\,t = 0,\,\,\,\,\,\,\,t = 2}}b Determine the intervals on which the function increases and decreases. Show Solution
To determine the increase/decrease information for the function all we need is a quick number line for the derivative. Here is the number line.
From this we get the following increasing/decreasing information for the function.
\require{bbox} \bbox[2pt,border:1px solid black]{{{\mbox{Increasing : }}\left( { - 3,0} \right)\,\,\,\,\& \,\,\,\left( {0,2} \right)\hspace{0.25in}\hspace{0.25in}{\mbox{Decreasing : }}\,\,\left( { - \infty , - 3} \right)\,\,\,\,\& \,\,\,\,\left( {2,\infty } \right)}}c Classify the critical points as relative maximums, relative minimums or neither. Show Solution
With the increasing/decreasing information from the previous step we can easily classify the critical points using the 1st derivative test. Here is classification of the functions critical points.
\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}t & = - 3 : {\mbox{ Relative Minimum}}\\ t & = 0 : {\mbox{Neither}}\\ t & = 2 : {\mbox{ Relative Maximum}}\end{align*}}