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