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(a) Find the intervals on which $ f $ is increasing or decreasing.(b) Find the local maximum and minimum values of $ f $.(c) Find the intervals of concavity and the inflection points.

$ f(x) = 2x^3 - 9x^2 + 12x - 3 $

a) the interval of increase: $(-\infty, 1),(2, \infty)$ and the interval of decrease: $(1,2)$b the local minimum is at $x=2 $ and minimum value is $f(2)=1$ and local maximum is at $x=1 $ and maximum value is $f(1)={2}$c) concave downward on $\left(-\infty, \frac{3}{2}\right)$ and concave upward on $\left(\frac{3}{2}, \infty\right).$ The inflection point is at $\left(\frac{3}{2}, \frac{3}{2}\right)$

05:44

Fahad P.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 3

How Derivatives Affect the Shape of a Graph

Derivatives

Differentiation

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given this function F of X. And the first thing to figure out the intervals where we're increasing or decreasing that we're going to have to do is find the derivative of this function. So f prime of X is equal to using the power rule eight X squared or sorry, not 86 X squared minus 18 X plus 12. And we can simplify this by taking a six out. So you get x squared minus three X plus two. And then we can factor this. So we get X -2 X minus one. And now we know that we have zeros zeros at X is equal to two and one since we factored this. And so what we're gonna do is we're gonna look at values of X that are less than one, look at values of X that are between one and two and then look at values of X that are greater than two. And we're gonna see what the sign of our derivative is. So if we plug in the value that's less than one, let's just say zero, we would get zero minus two is negative. Zero minus one is negative. So we get a negative times negative times positive, which is positive. So this is greater than zero. And now if you plug in the value of X it's between one and two let's just say 1.5, you get 1.5 -2 is negative. 1.5 -1 is positive. So we get a positive times negative times positive, which is negative, which means our Derivative is less than zero on this interval. And then the last interval that we want to look at is when X is greater than two. So let's just plug in three. We get three minus two which is positive three minus one, which is positive. So positive times positive times positive is positive. So are derivative is greater than zero on this interval as well. And now we know where our function is increasing or decreasing wherever are derivative of a function is greater than zero. Our function is increasing and wherever the derivative of the function is less than zero, our function is decreasing. So we're increasing from negative infinity to one or when X is less than one and from two to infinity Or an X is greater than two. And then we're decreasing in this interval from 1-2. So from 1-2 is where we are decreasing. And for part B, what we're gonna find are the local maximum minimum values. And the way that we do this is we have to find the zeros of our derivative, which we actually already find. We already found where X is equal to one and two in part A. And then we're also going to use use some information we found in part A. As well to figure out if these are maximum or minimum values. So if we look at values of X that are less than one Are derivative is positive. And then for values of X that are greater than one, but less than two are derivative is negative. So we're going from an increasing slope to a decreasing slope, which means that at X is equal to one. We have a local maximum whenever we have a zero at a point and we go from increasing to decreasing, we have a maximum if that zero is zero, the derivative of a function F. Fx. So the local max, he's at X is equal to one. And if we plug in one back into our function, we should get a value of To this is equal to two. So we have a max at the .1 common to. And now if we look at X is equal to two. Well our derivative was negative for values less than two, but greater than one. And then our derivative was positive for values greater than two. So we're going from decreasing to increasing slope or from positive to negative values of our derivative, which means this is a local minimum. So local men is at X is equal to two. And now we just plug back in two into our original equation and we should see that this is equal to one. So we have a local minimum at the .2 comma one. And now the last thing that we're gonna have to do in part C is figure out the intervals in which our function is actually concave upper, concave down and also find the inflection points. So if we look back at the equation for our derivative, it was six X squared minus 18 X plus 12. So the derivative of that would be 12 X. If I take this to down minus 18. So the second derivative of our function or EHF double prime of X, It's equal to 12 x -18. And so now what we wanna do is set this equal to 0 12, X minus 18 is equal to 0, 18 is equal to 12 X. And X is equal to 18/12, Which is equal to 3/2. So we know that we have a zero. The second derivative of a function is equal to zero at X is equal to 3/2. And now we just have to make sure that this is actually an inflection point. So what we want to do is look at values of X that are less than 3/2. And then look at values of X that are greater than 3/2. And see the sign for a second derivative. So if are the sign of our second derivative is positive. That means that we're concave up. And if the sign of our second river is negative, that means we're concave down. So If we plug in a value that of exits less than three halves, let's just say zero. We get 0 -18 which is negative. So our second derivative is negative on that interval. And if we plug in a value that is greater than three over to let's just say to you get 24 -18 which is positive. So her second relative is positive on this interval. So since we go from negative to positive, we're actually changing incan cavity from concave down to concave up, which means that this is indeed an inflection point at X is equal to three halfs. So what we can do to find the y coordinate of this point has just plug this into our original equation again. And if you do all these calculations and plug this back in, you'll see that this is equal to 3/2. So we have an inflection point at 3/2 comma 3/2. And now for the intervals in which we're concave up and concave down while we're concave down whenever our second derivative is less than zero. So we're concave down for all X being less than three halves. So you can say concave down from Negative Infinity to 3/2. And then we're concave up wherever our second derivative is positive, which was for all values of X that are greater than three half. So concave up From 3/2 to Infinity.

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