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Home / Calculus I / Applications of Derivatives / The Shape of a Graph, Part I
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Section 4.5 : The Shape of a Graph, Part I

4. This problem is about some function. All we know about the function is that it exists everywhere and we also know the information given below about the derivative of the function. Answer each of the following questions about this function.

  1. Identify the critical points of the function.
  2. Determine the intervals on which the function increases and decreases.
  3. Classify the critical points as relative maximums, relative minimums or neither.
\[\begin{array}{c}f'\left( { - 5} \right) = 0\,\,\,\,\,\,\,\,f'\left( { - 2} \right) = 0\,\,\,\,\,\,\,\,\,\,f'\left( 4 \right) = 0\,\,\,\,\,\,\,\,\,\,\,f'\left( 8 \right) = 0\\ f'\left( x \right) < 0\,\,\,\,\,{\rm{on}}\,\,\,\,\,\left( { - 5, - 2} \right),\,\,\,\left( { - 2,4} \right),\,\,\,\left( {8,\infty } \right)\hspace{0.25in}\hspace{0.25in}f'\left( x \right) > 0\,\,\,\,\,{\rm{on}}\,\,\,\,\,\left( { - \infty , - 5} \right),\,\,\,\left( {4,8} \right)\end{array}\]
Hint : This problem is actually quite simple. Just keep in mind how critical points are defined and how we can answer the last two parts from the derivative of the function.

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a Identify the critical points of the function. Show Solution

Okay, let’s recall the definition of a critical point. A critical point is any point in which the function exists and the derivative is either zero or doesn’t exist.

We are given that the function exists everywhere (and in fact this part is why that in there at all….) and so we don’t really need to worry about that part of the definition for this problem.

Also, from the given information about the derivative we can see that at every point the derivative is either zero, positive or negative. In other words, the derivative will exist at every point.

So, all this means that the critical points of the function are those points were the derivative is zero and we are given those in the information.

Therefore, the critical points of the function are,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 5,\,\,\,\,\,\,\,\,x = - 2,\,\,\,\,\,\,\,x = 4,\,\,\,\,\,\,\,\,\,x = 8}}\]

b Determine the intervals on which the function increases and decreases. Show Solution

There is really not a lot to this part. We know that the function will increase where the derivative is positive and it will decrease where the derivative is negative. This positive and negative information is clearly listed above in the given information so here are the increasing/decreasing intervals for this function.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{{\mbox{Increasing : }}\left( { - \infty , - 5} \right)\,\,\,\,\& \,\,\,\left( {4,8} \right)\hspace{0.25in}\hspace{0.25in}{\mbox{Decreasing : }}\,\,\left( { - 5, - 2} \right),\,\,\,\left( { - 2,4} \right)\,\,\,\& \,\,\,\left( {8,\infty } \right)}}\]

c Classify the critical points as relative maximums, relative minimums or neither. Show Solution

Okay, there isn’t a lot that we need to do here. We know that relative maximums are increasing on the left and decreasing on the right and relative minimums are decreasing on the left and increasing on the right. We have all the increasing/decreasing information from the second part so here is the answer to this part.

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}x & = - 5\,\,\,\,\,\,:\,{\mbox{ Relative Maximum}}\\ x & = - 2\,\,\,\,\,\,:\,{\mbox{ Neither}}\\ x & = 4\,\,\,\,\,\,\,\,\,:\,{\mbox{ Relative Minimum}}\\ x & = 8\,\,\,\,\,\,\,\,\,:\,{\mbox{ Relative Maximum}}\end{align*}}\]