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Numerade Educator



Problem 43 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) = x \sqrt{6 - x} $


a) increasing: $(-\infty, 4) \quad$ decreasing: $(4,6)$
b) local maximum: $(4,4 \sqrt{2})$
c) No inflection, Because the graph is concave down for all points in the domain.

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Video Transcript

and this problem, we are learning how the derivative or the 1st and 2nd derivative affect the shape of the graph. More specifically, we're going to be using the first and derivative test to determine things like the intervals of increase and decrease in the con cavity of a function. So for part A were given the function F X equals X time to square root of six minus X. So the first thing that we'll do is we'll take, um, the derivative, right. We're trying to find the intervals of increasing decrease. So your mind should be thinking I didn't take a derivative, So we'll take f Prime of X and we'll have to use the product will here and then once we simplify, the product rule will get f prime of X equals 12 minus three x all over to time to square root of six minus x. And then remember, we want the critical numbers. Now. This might be a little bit tricky because we have this fraction, as are derivative. Well, remember the that. If the numerator of the fraction is zero, then my whole fraction is zero, so we can set the numerator equal to zero to find the critical numbers, so we'll set zero equal to 12 minus three x will solve for X X is four. So now we confined the intervals of increase and decrease for our function. The domain of our function is from negative infinity to six, and our function does include six. So our first interval would be from negative. Infinity to four toe are critical number, and then what you could do is you could pick a number in this interval you could plug in to, for example, and then you can plug it into our function and determine the sign off that function. And when you do that, we'll find that it is positive. So it is increasing, and then our second intervals from 4 to 6 and you'll find that our function is decreasing there and now for be, we're told. Let's find the local maximum maximums and minimums well from the first derivative test and the sign of the function that we just found. We have a local maximum when X equals four so you can plug in for into the function to get F of four equals four time descriptive, too. So the point at which we have a local maximum is four comma, four square would tube. Now for part C, we're told to find the con cavity of the function. Now, when you hear the word con cavity, your mind should go to second derivative. We need the second derivative test to determine con cavity, so we'll take the second derivative. Now you might be looking at the first derivative and think that's a tricky derivative and you're right. You have to apply both the chain and the quotient rule. And then once you do that, f double prime of X equals three times X minus eight all over four times six minus X rays to the three halves. And then when we can see in this function is that our second derivative is less than zero for all X in our interval. Negative infinity to six. So that means that our function is con cave down when X is in that interval and now for D, we're told. Well, is our calculus correct? Can we compare it to a graph and say that what we found matched the graph of the function? And this is the graph of the function you can see that we got it. Correct. We have a local maximum here. We don't have any points of inflection are con cavity matches. So we did well. So I hope that this problem helped you understand a little bit more about how the derivative affect the shape of the graph and specifically how we can use the 1st and 2nd derivative to determine things like the interval of increasing decrease, local maximums and minimums and con cavity of a function.