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{displaymath}\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}\end{displaymath}    

      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{displaymath}\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}\end{displaymath}    

      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)$
      \includegraphics[scale=.7,angle=0,
clip= true]{q8-f0.eps}

      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.



Javier Fernandez 2004-03-24