Science:Math Exam Resources/Courses/MATH100/December 2011/Question 05 (e)/Solution 1

From parts (a) and (d), we know that the critical values for the first and second derivative of $f(x)$ are $x=-1,x=0$, and $x=4$. We first calculate the corresponding y-values for the x-values named above ...

• $\displaystyle x=-1$, $\displaystyle y=f(-1)=6(-1)^{1/5}+(-1)^{6/5}=-6+1=-5$
• $\displaystyle x=0$, $\displaystyle y=f(0)=6(0)^{1/5}+(0)^{6/5}=0$
• $\displaystyle x=4$, $\displaystyle y=f(4)=6(4)^{1/5}+(4)^{6/5}\approx 13.2$

...and then plot these points on the graph.

From part (b), we know the function is decreasing until the point $(-1,-5)$ and then increasing. Thus the function has a minimum at that point. Because the derivative is undefined at $(0,0)$, it has a vertical tangent line there.

Schematically, the tangent lines and increase/decrease of the function looks like this.

However, we know the function isn't made up of straight lines - it's more curvy than that. From part (d) we know that the function has an inflection point at $x=0$ and $x=4$. Furthermore we know that the function is concave up for $x<0$ and $x>4$ and is concave down for $0<x<4$. This tells us what our curves should look like.

If we superimpose the curve on top of the straight lines it looks something like this:

So the final graph should look like this: