Problem

(a) Find the intervals of increase or decrease. …

04:53

Problem 38 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) = 36x + 3x^2 - 2x^3 $

Answer

a) decreasing: $(-\infty,-2),(3, \infty) \quad$ increasing: $(-2,3)$
Decreasing on $[-2,2]$ $\\$
b) $\operatorname{Min}(-2,-44) \operatorname{Max}(3,81)$$\\$
c)Concave up: $\left(-\infty, \frac{1}{2}\right)$
Concave down: $\left(\frac{1}{2}, \infty\right)$
Inflection points: $x=\frac{1}{2}$$\\$
d) SEE GRAPH

Watch More Solved Questions in Chapter 4

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Video Transcript

and this problem, we are learning how derivatives affect the shape of a graph. Smart. Specifically, we're looking at the 1st and 2nd derivative to determine, um, intervals of increasing decrease and con cavity. Now you might be asking, Why do we learn this? Well, this tells us a lot of information about the behavior ever function, but also this is a precursor to curve sketching, which is a huge part of calculus. So this will help you understand curve sketching later on in your class. So for part A were told, Well, let's find the intervals of increase and decrease. The first thing that we need to do is find the derivative. So we're given ffx is negative two x cubed plus three x squared plus 36 x Well, we can simply use the power rule to differentiate it. So f prime of X is negative six X squared plus six x plus 36. Then, to find critical numbers, we set our derivative equal to zero and solve for X. So zero will be equivalent to negative six x squared plus six x plus 36 so we can factor out a negative six. Get night of six times the quantity X squared minus X minus six. And then this looks like a quadratic We know how to factor. So we get negative six times X minus three times X plus two and then we could just solve for X. Remember, this is equivalent to zero. So X would be negative two and X to be three, and then we can use this information to find the intervals of increase and decrease. The domain of dysfunction is native infinity to infinity. So all real numbers. So then what we're going to do is we're going to take the intervals, um, within our domain and then determine if our function is increasing or decreasing in that interval. So for the interval, negative infinity too negative to our function is decreasing. Now, if you are confused on how I determine that all you have to do is pick a number in the interval and plug it into the function and see if you get a positive or negative result. The second interval would be negative 2 to 3. We see an increasing function there and then three to infinity, and our function is decreasing there and now for part B, we're told. Let's find the maxes and men's of this function. What we solve for X X was negative 2 to 3. So let's just plug goes into our function. F of negative two is negative. 44 F of three is 81 so clearly we see a minimum at X equals negative two and a maximum at X equals three for part C, we're told. Let's determine Con cavity. Well, when you hear the word con cavity, your mind should immediately go to second derivative. We need the second derivative test to determine con cavity. So let's solve for F double Prime F double Prime of X is six minus 12 x. Again, we'll set that equal to zero to find the critical numbers. In this case, we only get one x equals one half. So then what we can do is we can pick two numbers within our interval and test further sign. So f double prime of zero equal six and that is greater than zero F. Double prime of one is negative six, and that's less than zero. So what does that mean? That means we see a con cave up behavior and from the interval negative infinity to one half and a con cave down behavior from one half to infinity. And then because we switch from con Cave up to con cave down, we have an inflection point at X equals one half and finally for D, it says, Let's check the graph of this function and see what we got was correct. So this is the graph of the function, and you consider that are intervals of con cavity, and increase and decrease are correct, and you can also go through this more specifically and determine the points as well. So I hope this problem helped you understand a little bit more about how the derivative can affect the shape of a graph. More specifically, how we use the 1st and 2nd derivative test to determine things like critical numbers, intervals of increase in decrease and the con cavity of the function

Madi S.

University of Denver

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 3

How Derivatives Affect the Shape of a Graph

Problem