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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) = x^3 - 3x^2 - 9x + 4 $

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(a) $f(x)=x^{3}-3 x^{2}-9 x+4 \Rightarrow f^{\prime}(x)=3 x^{2}-6 x-9=3\left(x^{2}-2 x-3\right)=3(x+1)(x-3)$(b) $f$ changes from increasing to decreasing at $x=-1$ and from decreasing to increasing at $x=3 .$ Thus, $f(-1)=9$ is alocal maximum value and $f(3)=-23$ is a local minimum value.(c) $f^{\prime \prime}(x)=6 x-6=6(x-1) . \quad f^{\prime \prime}(x)>0 \Leftrightarrow \quad x>1$ and $f^{\prime \prime}(x)<0 \Leftrightarrow x<1 .$ Thus, $f$ is concave upward on$(1, \infty)$ and concave downward on $(-\infty, 1) .$ There is an inflection point at (1,-7)

07:10

Fahad Paryani

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 3

How Derivatives Affect the Shape of a Graph

Derivatives

Differentiation

Volume

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University of Michigan - Ann Arbor

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to find the intervals where our function F of X is increasing or decreasing. The first thing we're gonna have to do is find the derivative of this function. So I'm just gonna have to use the power rule to find this derivative. The derivative of extra third is three, X squared minus six X -9. And then our constant goes to zero. So this is the derivative of our function. And now the easiest way, in my opinion to do this is to find the zeros of our derivative. So this is equal to three times X squared minus two x minus three, which is equal to three times if you factor this X plus one and x minus three. And now we can figure out the zeros of our derivative because we have factored this and they're gonna be at negative one and three. So F prime of X is equal to zero At X is equal to -1 and three. No. And now what we're gonna do is look at the intervals where X is less than negative one, where X is between negative one and three And where X is greater than three. And we're going to see the sign. We're gonna look at the sign of our derivatives, so for X is less than negative one. If we plug in a value let's just say negative to negative two plus one is a negative value and the negative two minus three is also negative. So we have a negative times a negative times positive which is positive. So this is greater than zero. And then from -1, 23 if we pick a value of X equal to 00 plus one is 10 minus three is negative three. So we have a negative times positive times positive, so are derivative is negative in this interval. And then the last one is when X is greater than three, let's just pick 585 plus one is six times five minus three, which is two. So these are all positive values. So again are derivative is positive on this interval and whenever a derivative is positive, that means our function is increasing and whenever are derivative is negative, that means our function is decreasing. So we're increasing from negative infinity to negative one and from 3 to Infinity. Since we found that X values less than negative one, we have positive derivative of F of X. And for X values greater than three, we also have a positive value for derivative Fx. So that's where we are increasing and then we're decreasing. Is this interval from negative 123 So we're decreasing from negative 123 Now, for part B we want to figure out the local minimum and maximum values of our function F. Of X. And these occur when are derivative F. Prime of X. It's equal to zero. So what we're gonna want to look at is the values of X is equal to negative one and 23 Since we found that F. Prime should be F. Prime, not ffx um Is zero at these points. And so what we're gonna do is we're going to look at the actual data that we already found. So Gonna look at values of X that's less than -1. You see that are derivative is positive and the values of X between negative one and three are negative. So we're going from positive to negative, which means we're going from an increasing slope to a decreasing slope, which means that at X is equal to negative one, we have a local maximum. So if we plug this back into our original function, see that f of negative one Is equal to nine. So we have a local max at nine. And now for our other value at X is equal to three. If we look at values that are between negative one and three, that gives us negative values for our derivative of F. Of X. And if we look at values that are greater than three, we get positive. So we're going from negative to positive, decreasing slope to increasing slope, which means that this is a local minimum. And if we plug this into our original function, We'll see that this is equal to -23. So we have a local maximum at the point negative 19 and a local minimum at the 190.3 negative 23. And now for part C we're going to figure out the intervals in which our function is concave up or concave down and then also where our inflection points are. So the way that we're gonna do this is by finding the second derivative of our function f double prime of X. And if we look at F prime of X again in its original form it was three x squared minus six x minus nine. So the derivative of that, it's going to be six x minus six. So our second derivative is equal to six X -6. And now we want to figure out where this is equal to zero. So I'm just going to factor out of 66 times X -1. So at X is equal to one, X is equal to one, Our second derivative of X is equal to zero. So we know that there's this point at X is equal to one where our second derivative is equal to zero. And wherever our second derivative is negative are function is going to be concave down and wherever our second river is positive or function is going to be concave up. So what we want to do is look at values of X that are less than zero and see what the sign of our second derivative of X is. So if we plug in a value, let's just say of negative one, we have negative one minus one which is negative to my time six, which is a negative value. So these Are going to be values that are less than zero. And if we look at X values that are greater than zero, Let's just plug in one. Sorry, I should not have been looking at X is less than zero and X is greater. This should be X is less than one. X is greater than one. So if we plugged in a value of zero we would have a negative one times six which is still negative. So yes, we do have our second derivative being less than zero and X is less than one. And when X is greater than one, if we plug into value, let's just say if two, We have 2 -1, which is one times six is six, which is positive, so are derivative. Our second derivative of F of X is positive when X is greater than one. So this means that we are concave down on the interval from negative infinity to one, since for values of X less than one are second derivative, F double prime of X is less than zero. And we're concave up On the interval from 1 to infinity, since our second derivative is positive on this interval. And now, lastly, the inflection point, these points are where we go from concave up to concave down or from concave down to concave up. So whenever we have a change in common cavity, we have an inflection point and these occur when our second derivative is equal to zero and we have a sign change. So if we were to look at the graph of our second derivative f double prime of X, we would need a zero that actually crosses the X axis. We couldn't have a zero that just touches and then goes back like this. Since that doesn't have a sign change, which means we are going from concave in this case we're going from negative to negative. So we're going from concave down to concave down, which isn't a change in common cavity, which means it is at an inflection point. So we know that from finding values on these intervals that X is equal, two values that are less than one are derivative is negative in X is equal to values that are greater than one are derivative is positive. So we do have a sign change at X is equal to one, which means that our inflection point is at X is equal to one, and that's our only inflection point.

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