Science:Math Exam Resources/Courses/MATH100/December 2012/Question 06 (f)/Solution 1

Before assembling the graph, we will calculate the x-intercepts, found by setting the original function equal to zero and solving for x.

$2x^{3/5}+3x^{-2/5}=0$

Factoring out $x^{-2/5}$ gives:

$x^{-2/5}(2x+3)=0$

So there is an x-intercept at $x=-3/2$

To draw the graph, we will start by plotting the vertical asymptote at x = 0 and various points: the x-intercept at $(-3/2,0)$, the minimum at $x=1$ and the inflection point at $x=7/2$. Because the question asks for the y-coordinate of the minimum (from part b), we plug x = 1 into f(x) to get:

$f(1)=2(1)^{3/5}+3(1)^{-2/5}=2+3=5$

So our minimum will be the point $(1,5)$. All this gives a picture that looks like this:

Using information from part a) about the intervals of increase and decrease, we know the graph should roughly look like this:

Finally, using part c), we know the function is concave up for $x<7/2$ and concave down otherwise, which leads us to the final graph: