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Problem 39 Medium Difficulty

(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.

$ f(x) = \frac{1}{2} x^4 - 4x^2 + 3 $

Answer

(a) $f(x)=\frac{1}{2} x^{4}-4 x^{2}+3 \Rightarrow f^{\prime}(x)=2 x^{3}-8 x=2 x\left(x^{2}-4\right)=2 x(x+2)(x-2) . \quad f^{\prime}(x)>0 \Leftrightarrow-2<x<0$
or $x>2,$ and $f^{\prime}(x)<0 \Leftrightarrow x<-2$ or $0<x<2 .$ So $f$ is increasing on (-2,0) and $(2, \infty)$ and $f$ is decreasing on
$(-\infty,-2)$ and (0,2)
(b) $f$ changes from increasing to decreasing at $x=0,$ so $f(0)=3$ is a local maximum value.
$f$ changes from decreasing to increasing at $x=\pm 2,$ so $f(\pm 2)=-5$ is a local minimum value.
(c) $f^{\prime \prime}(x)=6 x^{2}-8=6\left(x^{2}-\frac{4}{3}\right)=6\left(x+\frac{2}{\sqrt{3}}\right)\left(x-\frac{2}{\sqrt{3}}\right)$
$f^{\prime \prime}(x)=0 \Leftrightarrow x=\pm \frac{2}{\sqrt{3}}, \quad f^{\prime \prime}(x)>0$ on $\left(-\infty,-\frac{2}{\sqrt{3}}\right)$ and $\left(\frac{2}{\sqrt{3}}, \infty\right)$
and $f^{\prime \prime}(x)<0$ on $\left(-\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\right) \cdot$ So $f$ is $\mathrm{CU}$ on $\left(-\infty,-\frac{2}{\sqrt{3}}\right)$ and
$\left(\frac{2}{\sqrt{3}}, \infty\right),$ and $f$ is $\mathrm{CD}$ on $\left(-\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\right) .$ There are inflection points at
$\left(\pm \frac{2}{\sqrt{3}},-\frac{13}{9}\right)$

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Video Transcript

and this problem, we are learning how derivatives affect the shape of a graph. More specifically, this problem is asking us how we can use the derivative to deduce important information about the function, like where the function is increasing or decreasing three intervals of con cavity. And it's critical numbers. And this is going to become important later on calculus when you start to learn curve sketching, so getting a great understanding of this now is going to be very helpful for you. Sorry. Pardon me for part A were given the function F of X equals in one half X to the fourth minus four X squared plus three. So we need to find the critical numbers and the intervals of con cavity. So the first thing that we're going to do is take the derivative F prime of X equals two X cubed minus eight x We could do some. Factoring will get two x times X minus two times X plus two. We're going to set that equal to zero to get the critical numbers. Then we'll solve for X so x zero X s two and excess negative too. So then what? We're going to Dio is we're going to create a table. You don't have to make a table, but just keep track of the intervals. We're going to make sure that we have the intervals using the critical numbers and then determine if our function would be increasing or decreasing in that interval. So the first interval is negative. Infinity too negative too. And our function is decreasing there. And if you don't know how I'm getting that, all you have to do is take a number within that interval, plug it into the function and see if you get a positive or a negative. For the second interval, we'd have negative 2 to 0 and we see our function is increasing there. Our third interval interval. Pardon me is from 0 to 2 and our function is decreasing there. And then finally, our last interval is to to infinity and our function is increasing there. So this is just another way to right. We found on the table at the vex increasing when X is an interval. Negative 20 and two comma infinity and f of X is decreasing on the interval when X is in thes two intervals negative and they needed to negative negative two or two or pardon me, you 02 and then for party were told. What are the maxes and men's of this function? What we have the critical numbers. All we have to do now is plugged them back into our original function. So f of negative two is negative. Five f of two was negative. Five and F of zero is three. So clearly we have a minimum, a negative two and two and a maximum at zero for part C, we're told. Let's check the con cavity. So when you hear the word con cavity your mind, you immediately go to second derivative. We need the second derivative test to determine the intervals of con Cavity. So F double prime of X is six x squared minus eight. And now this one's a little bit more confusing than before. We have to do some clever factoring. So this is the same thing as saying six times X squared, minus 4/3. And then we want critical numbers, Remember? So we have to factor this further. We're going to get six times X minus two over the square root of three times X plus two over the square root of three. So then we have our critical numbers so we can determine con cavity. Our function is con cave up on the interval. Negative infinity negative to over screwed of three and two over. Screwed of three common infinity And then we see a con cave down behavior on the interval negative to over the square root of 3 to 2 over the square root of three. So what does this mean? We change from going from concave up to con cave down So we see an inflection point when x is negative two over the skirt of three and when X is two over the square root of three. And now finally the problem asks us, Well, what's going to happen with the graph? Can we check our information from the graph? This is the graph of the function and you can clearly see that our information that we found using just differential calculus matches the function. We see our minimums where we found them, our maximum where we found them. Our con cavity matches as well. So I hope this problem helped to understand a little bit more about how the derivative, um, affect the shape of the graph more specifically, how we can use the derivative to deduce important information from a function such as the intervals of increase in decrease and it's common cavity.