Section 4.5 : The Shape of a Graph, Part I
5. For \(f\left( x \right) = 2{x^3} - 9{x^2} - 60x\) 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.
\[f'\left( x \right) = 6{x^2} - 18x - 60 = 6\left( {{x^2} - 3x - 10} \right) = 6\left( {x - 5} \right)\left( {x + 2} \right)\]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,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 2,\,\,\,\,\,\,\,\,x = 5}}\]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.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{{\mbox{Increasing : }}\left( { - \infty , - 2} \right)\,\,\,\,\& \,\,\,\left( {5,\infty } \right)\hspace{0.25in}\hspace{0.25in}{\mbox{Decreasing : }}\,\,\left( { - 2,5} \right)}}\]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.
\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}x & = - 2 : {\mbox{ Relative Maximum}}\\ x & = 5:{\mbox{ Relative Minimum}}\end{align*}}\]