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(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 inforvation from parts (a)-(c) to sketch the graph. Check your work with a graphing device if you have one.$h(x)=5 x^{3}-3 x^{5}$
A. $h(x)$ is increasing on $(-1,0) \cup(0,1)$$h(x)$ is decreasing on $(-\infty,-1) \cup(1, \infty)$B. Local minima at $x=-1$ and maxima at $x=1$C. $x=-\frac{1}{12}$, $x=0$, $x=\frac{1}{\sqrt{2}}$D. SEE GRAPH
Calculus 1 / AB
Chapter 4
APPLICATIONS OF DIFFERENTIATION
Section 3
Derivatives and the Shapes of Graphs
Derivatives
Differentiation
Applications of the Derivative
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for this question were first right out to the dearer to yourself if for part one we said, I've prompted the zero. So you have X equals zero minus during the one and we can see that the four hour intervals next if infinity to a negative one negative 10 0 to 1 in the one to infinity on the first interval, we can see that if exists, that's though that's no one. So if crime is negative, the phone is decreasing. On the second her phone negative 1 to 0 If promise qualitative. The function is increasing on the sick on the third interval. If if X is below 0 to 1 if prime is positive, the finishing secrecy If access is great on the what if crime is connective The funny on the Funchess decreasing. So we have a local Oh minima at X equals two minus one with with value F minus one in close to minus two and then we have a local next Men at X equals to one with value if one because the to So this is part true. Have three. We set a second narrative to be zero, so we have X equals zero. Um, I and the process minus one over wrote off to. So we consider the following for intervals minus infinity to minus one over to talk to minus one over it up to two, zero 0 to 1. Over what? Of two at the one over. Root of two to infinity. So we can see the sign of this. So no first interval, we can see that f double promise. Positive. So the functions conflict on the second ago. We can see that if double prime just directive so the functions can keep down on the third interval. If the book crime Yes, positive. So the functions can keep up. And on the last interval of stop a promise next trip, the functions can't get them in the Based on this information, we are ready to schedule a graph off. If so, first we laboard's on the local minima local maximum. Um, it's X equals two minus on the X seaQuest one. So we have minus one in the one said. Of course, bone being bad, it will be minus two and to Okay, so for looking for the inflection point, we have to re inflection point at X equals two finals one over. Ripped off too. And the X equals zero. And that cakes equals to 1/2. Okay, so we are ready to graph. Yeah, relate. So really boards the inflection point by the green. Common. Okay. It's a very the graph. So the function is decreasing. First in the conclave up and the increasing Now you become concave down. It did not become Come caper up in the conclave. Dumb again. No decrease in. So this is the grab off function. If so, we have three inflection point. And that one no come makes him and the one no convenient.
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