Graph each function. Determine the largest open intervals of the ...

Question

Graph each function. Determine the largest open intervals of the domain over which each function is (a) increasing or (b) decreasing. See Example 1. ƒ(x)=1/3(x+3)^4-3

Accepted Answer

Hey, everyone in this problem for the following function, we're asked to graph and to identify its domains open intervals for which the function is increasing or decreasing. The function we're given is F of X is equal to 1/ multiplied by X plus five to the exponent four minus five. We're given four answer choices, options A through D A and B have the same graph shown for the function. But they differ in the intervals where the function is increasing or decreasing. And the same goes for option C and D, they show the same graph for the function but have different intervals for the decreasing and increasing portion. So let's go down and give ourselves some space to work out this problem. We're just gonna rewrite the function we were given, OK. We have the function F of X is equal to 1/5 multiplied by X plus five to the exponent four minus five. And the first thing we're asked to do is to graph this function. So if we want to graph this function and we can see that there are several transformations applied. We want to think about what the parent function is. OK. So in this case, we're gonna call it G FX in Geva equals X to the exponent four is the parent function. So we start with G FX equals X to the exponent or we apply a bunch of transformations and we end up with our function F of X. We're gonna walk through those transformations one by one to see how they change the vertex of this. OK. All right. So for geo the vertex is located at the origin at 00. OK. For X to the exponent four, we know that if X is equal to zero G of X is gonna be equal to zero. This function opens up and that's gonna be that minimum point right at that origin. So let's start with our vertical compression. OK. F of X is going to be vertically compressed. That's what that multiplying by 1/5 tells us. Now, if we vertically compress this function, OK? It originally had its vertex at 00. That's actually not gonna change the location of its vertex. Vertex does not change in this case. Now, we move to the next transformation. The next transformation is gonna be that F of X is shifted downwards five minutes. Then let me go ahead and color code these transformations. OK. So the first one is going to be red and that's indicated by that 1/5. And this next one we're talking about, I'm gonna do this one in blue. This is that shifted down and that's that minus five at the end. OK. We take the function, we do some things to it and then we subtract five. And that is gonna shift that graph downwards five units. OK. So if we shift the entire graph downwards five units, that means we're gonna be shifting that vertex downwards five units. This is going to affect the Y coordinate and our vertex is going to be at zero comma negative five. OK. So we've done our two first two transformations, we have one final transformation to worry about. We're gonna talk about this one in green. And this is the plus five inside the brackets. We have X plus five to the exponent four instead of just X to the exponent four. If we add five just to that X term inside of the bracket, this is going to be a horizontal shift and it is going to shift this graph to the left five minutes. All right. So our graphic gets shifted to the left five units. What happens to the vertex again, this is a horizontal shift. It's affecting the X coordinate. So our X point is now going to be at negative five. So our vertex is gonna be located at negative five common whoops, negative five. All right. So we have the vertex of our problem located at negative five comma negative five. We know that this is a even function or an even exponent function, sorry even exponent polynomial that opens upwards. If we compare this to our answer choices right now, we can see that we can already eliminate option A and B because they have the vertex in the wrong location. OK. They have the vertex at positive five and negative five instead of negative five and negative five. OK. So we've eliminated A and B based off of the graph. OK. We know that the graph in option C and D is correct, the vertex is in the right location. The function is opening upwards like it should. OK? And now we need to figure out what happens on the left and the right of that vertex to see when the function is increasing and decreasing. If we look at this function and we're gonna look at the graph and we wanna figure out where it's increasing and where it's decreasing and well to the left of the vertex, this function is decreasing. It's going down, down, down, down, down until it hits the vertex. And then it's gonna go up, up, up, up, up, OK. So it's gonna be decreasing from negative infinity all the way to negative five, from negative five to positive infinity. It's going to be increasing. OK? And so C is actually going to be the correct answer here. It has the correct intervals for increasing and decreasing. If we look at option D, it says that it's only decreasing from negative eight to negative five and from negative five to negative two. OK. Now these are the intervals that are shown on the graph but that doesn't mean that's the entire interval for our function. OK. This function continues beyond this graph. And so we have to consider those points as well. So the correct answer here is option C we have A X to the exponent four graph that has been transformed. The vertex is located at negative five comma five. It opens upwards, we have this function decreasing from negative infinity to negative five and increasing from negative five to positive infinity. Thanks everyone for watching. I hope this video helped see you in the next one.

Struggling with College Algebra?

Graph each function. Determine the largest open intervals of the domain over which each function is (a) increasing or (b) decreasing. See Example 1. ƒ(x)=1/3(x+3)^4-3

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