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MAC2311 #10122Test: Test 3 This Question: 4 pts15 0f 25 (1 complete) -Find the intervals on which f is increasing and the intervals on which it is decreasing: f(x) ...

Question

MAC2311 #10122Test: Test 3 This Question: 4 pts15 0f 25 (1 complete) -Find the intervals on which f is increasing and the intervals on which it is decreasing: f(x) 3x3 + 9x2Select the correct choice below and, if necessary fil in the answer box(es) to complete your choice. The function is decreasing The function is never increasing (Simplify your answer Use comma t0 separate answers as needed. Type your answer in Interval notation ) The function is decreasing on and increasing on (Simplify your

MAC2311 #10122 Test: Test 3 This Question: 4 pts 15 0f 25 (1 complete) - Find the intervals on which f is increasing and the intervals on which it is decreasing: f(x) 3x3 + 9x2 Select the correct choice below and, if necessary fil in the answer box(es) to complete your choice. The function is decreasing The function is never increasing (Simplify your answer Use comma t0 separate answers as needed. Type your answer in Interval notation ) The function is decreasing on and increasing on (Simplify your answers Use comma t0 separate answers as needed Type our answers in interval notation ) The function is increasing on The function is never decreasing (Simplify our answer Use comma to separate answers a5 needed Type your answer in interval notation ) The function is never increasing or decreasing Click t0 seleci ana enter your answeris;



Answers

Use the graph of the function $f$ to answer the following questions. On what interval(s) is the function $f$ decreasing?

Right. We want to identify with the function of increasing or decreasing. Activex is equal to X ray to the power of 3/5 as the F prime of X symbol with the Astros from the left. To organize the first derivative solve this problem. So the first derivative of F of X is 3 50 X negative two fists for 3/5. Excellently fits here. Remember that We can use the first ever to solve this problem by realizing that the function F is increasing whenever F prime of X is greater than zero and F of X is decreasing whenever a private X is less than zero. So to figure out the integral where we're going to test for the sign of at the time of X, we're gonna look for a santos 40 in the derivative. So there's no zeros but there is an essential X equals zero. So we check extra left into the right of zero. For excellent, negative. 10 and 10 for instance, we have the F Prime is positive and positive, since X equals zero is defined for the function F is continuous at zero. We can simply write that. It is increasing them for all X. Because of increasing both the left and right of zero and zero is defined. So that means that the function F is increasing for all of X and decreasing nowhere.

We want to know where this function is. Increasing, decreasing It is a polynomial function. So when we examine the critical points, we know that it can't be non differential. So we really just want to look for critical points where the derivative is zero. The derivative will involve the power rule derivative of two X to the fifth would be 10 x to the fourth for the derivative of 15 x of the fourth over four. I'm gonna bring that for in front and subtract one from its power. So the four that I bring in front actually cancels with the denominator, and then the power drops to the third power and then the last term. If I bring the three in front, it will cancel with the Dominator, and my new power is just X squared. Critical points will be where this is equal to zero. So let's set 10 acts of the fourth minus 15 x cubed plus five x squared equal to zero. We can pull a common five x squared outfront, which would give me two X squared minus three x plus one. Okay, we knew a little trial and air on the two x squared minus three x plus one. But that will factor 22 acts minus one times X minus one. Now, any time that we have that equal to zero, we've got a critical point. Well, five X squared is equal to zero when X equals zero X minus one is equal to zero when X equals one for the two X minus one. If I set that equal to zero, it's really just a two step equation. I have one divide by the two, so I get an answer of X equals 1/2. Okay, so now we're going to analyze this. We're gonna put this on a number line and check the sign of our derivative be of X equals zero X equals 1/2 or 0.5 X equals one. Now we will pick points on all sides. Let's start with the X value of negative one, and we're filling this into F prime. If I feel that in for F prime, I get 10 plus 15 plus five, definitely a positive. So we're looking at an increasing function now. We need to be ah, little bit careful with these regions that are coming up because we have them as decimals. Let's try filling in 0.25 for the value between 0.5. So I'm gonna go 10 times, 100.25 to the fourth, minus 15 times 150.25 to the third, plus five times 50.25 square. That gives me a really small, positive answer. I get 0.117 so it's still increasing for a value between 0.5 and one. I'm going to use 10.75 if I type that in 10 times 0.75 to the fourth, minus 15 times 150.75 cubed plus five times 50.75 squared. I get a slightly negative answer, so that's a decreasing region. Finally, if I fill in two 10 times, two to the fourth, minus 15 times two cubed plus five times to square in answer of 60. So that is a positive derivatives, so it increases, so there's actually two increasing regions. This whole region here is considered increasing. Zero is increasing because it didn't ever change to a downward trajectory. So we're looking at an increasing function, from negative infinity to 1/2 as well as one to infinity. The Onley decreasing region is between the X values of 1/2 and one

All right. We want to identify where the function is increasing or decreasing function in question is F of X equals x minus five. All raised to the power of two thirds as the F prime of X symbol. With the Astros here on the left hand set, we're going to use the first derivative to solve this problem. Remember that The first derivative can be used to identify where function increasing or decreasing namely after the X is increasing or F prime of X is greater than zero and decreasing the F prime of X is less than zero. So we have the F prime of X. S written on the left here is two thirds one over cube root X minus five. So we look for a scapegoat for zeros in the prime of X to determine what intervals of community test. There are no zero. So we have an acid coated X equals five. So we test the left and right of five. So for instance in the chest X equals zero and X equals 10. We find the F prime is negative left of five and positive right of five, which means that we're increasing for X greater than five and decreasing for X less than five for Fx.

We want to identify with the function is increasing or decreasing. F of X is equal to X to the power of two thirds plus five as the F prime of X symbol to left with the asterix. Hence that we're gonna use the first derivative to solve this problem. Remember that as I've written on the right, the first derivative can be used to find more functions increasing and decreasing. By noting that f is increasing for f prime of X zero and decreasing where it's derivative is less than zero. So F prime of X is as I've written here on the left to over +33 group X. So we can look for ascent oats in zeros of F prime to determine where we want to split our interval into or rather what areas of experts attest to sign up for the first derivative. So we don't have any zeros, but we haven't asked X equals zero. So, however, we do know that Quebec's is continuous at X equals +02 Essential. So for the accident, X equals zero, which has led to the left and right. So for X equals negative. 10 and 10. For instance, F prime is negative and positive, which means that we are increasing for X greater than zero, decreasing for X less than zero.


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