Quiz 8 (Solution)

Date: MATH 1210-4 - Spring 2004

Note: Unless stated otherwise, answers without justification receive no credit. Please, show your work!

1. Let $g(x)=3x^5-5x^3+1$
   1. Find all the critical points of $g(x)$.

      We have $g'(x) = 15x^4-15x^2 = 15x^2(x^2-1)$. Then, the stationary points are $g'(x)=0 \Leftrightarrow 15x^2(x^2-1) = 0$, but $15x^2(x^2-1)$ vanishes only if $x=0$ or $x^2=1$. Then, the stationary points are $0,-1,1$.

      Since there are no endpoints nor singular points, the only critical points are $0,-1,1$.

   2. Find the regions where $g(x)$ is increasing and where it is decreasing.

      We find this by analyzing the sign of $g'(x)$. Since $g'$ vanishes at $-1,0,1$, it suffices to evaluate

      $$\begin{split}g'(-2) & = 15\cdot 4\cdot 3 >0\\ g'(-\frac{1}{2}... ...\frac{1}{4} - 1) <0\\ g'(2) & = 15\cdot 4\cdot 3 >0 \end{split}$$

      
      
      to conclude that $g$ is increasing on $(-\infty,-1)$ and on $(1,+\infty)$, while it is decreasing on $(-1,1)$.

   3. Find the regions where $g(x)$ is concave up and where it is concave down.

      We find the concavity of $g$ by studying the sign of $g''(x) = 60x^3-30x = 30 x(2x^2-1)$. As usual, we start by finding where $g''(x)=0$, that is where $30 x(2x^2-1)$ vanishes, which only happens if $x=0$ or $2x^2=1 \Leftrightarrow x =\pm \sqrt{\frac{1}{2}}\sim \pm 0.707$. Then, it suffices to evaluate

      $$\begin{split}g''(-1) &= 30\cdot(-1)\cdot(2-1) <0\\ g''(-0.1) ... ...\cdot 0.01 -1)<0\\ g''(1) &= 30\cdot 1\cdot(2-1) >0 \end{split}$$

      
      
      to conclude that $g$ is concave up over $(\frac{-1}{\sqrt{2}},0)$ and $(\frac{1}{\sqrt{2}},+\infty)$ while it is concave down over $(-\infty,\frac{-1}{\sqrt{2}})$ and $(0,\frac{1}{\sqrt{2}})$.

   4. Find all inflection points of $g(x)$.

      An inflection point is a point where the concavity of $g$ changes, and this happens at $x=\frac{-1}{\sqrt{2}}$, $x=0$ and $x=\frac{1}{\sqrt{2}}$. These are the three inflection points of $g$.

   5. Using the first or second derivatives of $g(x)$ decide which of the critical points are local maximum values and which are local minimum values.

      Using the first derivative we see that, coming from the left, $g$ changes from increasing to decreasing at $x=-1$, so it is a local maximum; at $x=0$ $g$ continues decreasing (so $g$ doesn't have an extremum value there), and at $x=1$ $g$ goes from decreasing to increasing so that at $x=1$ $g$ has a local minimum.

      The local maximum value is $g(-1)=3$ and the local minimum value is $g(1)=-1$.

      Alternatively, the same results for $x=-1$ and $x=1$ are obtained from $g''(-1) = -30 < 0$ and $g''(1)=30>0$. But, the second derivative test doesn't decide the behavior of $g$ at $x=0$ since $g''(0) = 0$.

   6. Using only the information that you have gathered so far, sketch a graph of $g(x)$ showing this information.

      Figure 1: Graph of $g(x)$

      

      Figure 1 shows the sketch, where the critical and inflection points are marked. Notice that the increasing and decreasing regions, as well as the concavity matches our results from the previous items.