💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

Numerade Educator

# (a) Find the intervals of increase or decrease.(b) Find the local maximum and minimum values.(c) Find the intervals of concavity and the inflection points.(d) Use the information from parts $(a) - (c)$ to sketch the graph. Check your work with a graphing device if you have one.$G(x) = 5x^{2/3} - 2x^{5/3}$

## a) decreasing: $(-\infty, 0),(1, \infty) \quad$ increasing: $(0,1)$b) local minimum $G(0)=0 \quad$ local maximum $G(1)=3$c)$\mathrm{CU} :(-\infty,-0.5) \quad \mathrm{CD} :(-0.5,0),(0, \infty) \quad$ inflection point $\approx(-0.5,3.78)$d) SEE GRAPH

Derivatives

Differentiation

Volume

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

derivatives and second derivatives are very powerful. Ultimately, allow us to, um, determine the shapes of graphs. That's one really special use of them. And that's what we're going to see in this problem. The fact that we can take derivative, we could take the second derivative and were able thio see minimums and maximums, con cavity and a lot of other really important details within graphs. Okay, so this is the graph that we're dealing with for this problem in particular. Um, and we first want to determine intervals of increase and decrease. So we see the graph is decreasing from negative infinity to zero. It increases from 0 to 1 and then decreases from one to infinity. Then we want to determine local minimums and maximums. We see that there's a local minimum right here in 00 I'm Interestingly, if we look at the derivative craft, we see that, um, it's undefined at this 0.0, which also indicates that it's critical point on a local minimum. Then we see there's a local maximum at one, and it's this value here 13 With that, we now want to move on to Concha Vitti. We see that the graph is con cave up until negative one half and then it becomes con cave, um down and it's undefined zero. Then it goes con cave down. Um, it remains con cave down for the rest of the graph. I mean, when we have the sharp turns and make things really interesting because the graph is not actually differential at that point, then we have, um Now, since we've determined con cavity, we can also determine inflection points. We see that this is where the second derivative graph equals. Zero spot tells us that, um, negative 0.5 is an inflection point of this graph.

California Baptist University

Derivatives

Differentiation

Volume

Lectures

Join Bootcamp