Graphing functions Use the guidelines of this section to make ...

Question

Graphing functions Use the guidelines of this section to make a complete graph of f.

f(x) = ln (x² + 1)

Accepted Answer

Welcome back, everyone. In this problem, we want to draw the graph of the function F X equals the natural log of 3 X2 + 5. Now, if we're going to draw this graph, there are a few things that we need to know. First, we need to know the functions domain, OK? Now what is the domain of the natural log of 3 x2 + 5? Well, by definition, the natural log is only defined for positive values of its argument. In other words, 3 x2 + 5 would have to be greater than 0. No, because we have a Xquared term here that means whatever our X term is, whether it's positive or negative, when we square it, it's gonna become positive, OK, which tells us then that the domain. Let me put it in red here. The domain is gonna go from negative infinity to positive infinity. It's going to include all real numbers, OK? Because 3x2 + 5 is always positive for all real X. Next, we need to find the functions symmetry, OK? And by symmetry, that means we're trying to figure out if the function is either odd or even, or neither. Now, let's start by checking if it's odd. If it's even, sorry. A function is even, OK? If F of negative X is equal to F X. Now is that true for our function? Well, F of negative X, OK, is going to be equal to the natural log of 3 multiplied by negative X squared, OK, plus 5. And that would work out to the natural log of 3 X2 + 5, which is F of X. Therefore, this tells us that our function is even. And because it is even, that means FF X is symmetric. About the Y axis. Now what else do we need to know? Well, OK, another thing we need to do is to find the critical points of our function, OK? And at the critical point, OK, at the critical point or critical points. OK. Let me just fix that here at the critical points. Then the first derivative of our function would be equal to 0. What's the first derivative of F of X? Well, if we differentiate it. OK, to differentiate a natural log, we're going to write the argument in the denominator and the derivative of the argument in the numerator, and its derivative is 6X. Therefore, the first derivative is 6 X divided by 3 X2 + 5, and if that is equal to 0, then X must be zero. That's the value that's going to make the first derivative 0. So this tells us that there is a critical point at X equals 0. No, we can also determine increasing or decreasing intervals here because when X, if we consider it at our critical point, when X is positive, when it's greater than 0, then F is increasing, OK. And when X is negative, when it's less than 0, F is decreasing, OK. That's also something that we can keep in mind. So we have our critical point. We know the graphs symmetry, we know its domain. Next, we need to think about the graphs inflection points, OK? Now, at the inflection point, OK, at the inflection point. OK, we know that the 2nd derivative of our of our function will be equal to zero. And now it's easier for us to find the second derivative because we already know the first derivative. So we're going to differentiate our first derivative with respect to X. Notice that our first derivative is a quotient, so we can use the quotient rule of differentiation. And when we do that, OK, I'll leave you to work the details in the interest of time, but when we do that in our numerator, we're gonna get 6 multiplied by 3 X2 + 5. The derivative of the numerator multiplied by the denominator minus 6 X multiplied by 6 X, derivative of the denominator multiplied by the numerator, all divided by the numerator squared. 3 x 2 + 5, all squared. OK? So this is the second derivative of F of X. Now when we simplify our numerator here, we're gonna end up with 18 X squared. Plus 30, OK. All divided by 3 X + 5 are squared. This is the second derivative. Now, using what we know about inflection points, let me just make some space here. OK. Then, 8-18 X squared. Plus 30 divided by 3 X plus 5 all squad will be equal to 0 at the inflection point. And if that's the case, if we multiply through by +33 x squared should be 3 x 2 plus 5 r squared, oops I forgot it down here as well, my apologies. Then we're going to get 18 X squared plus 30 to be equal to 0 and know if that's the case, OK. Then that means 18 x 2 is going to be equal to 30. X squared equals 30 divided by 18 or 5/3, so that means X will be equal to plus or minus the square root of 5/3. So this is the X value at our inflection point. Now, what about the graphs con concavity? Let me just pronounce that properly here. Well, This tells us then that if is concave up, OK. Concave up for the absolute value of X to be greater than the square of 5/3, and F is concave down. OK, for the absolute value of X to be less than. Sorry, the other way around. It should be concave up for the absolute value of X less than 5/3 and concave up concave down for the absolute value of X greater than 5/3, OK? So we found the inflection point and the concavity, OK? Now, what about our intercepts? But let's put that as point number 5. What about our intercepts? Well. OK. Or are intercepts. OK, then if we think about it here for the Y intercept, that is when X is going to be 0, so that means we're going to be finding the Y value F of 0, which will be equal to the natural log of 3 multiplied by 02 + 5, OK, which will be equal to the natural log of 5. So this tells us that the Y intercept, OK, which is also the critical point is going to be LN5. So the point is going to be 0 LN5 and this point is the local minimum, OK? So keep that in mind here. This point is our local minimum. Now when it comes to our X intercepts. OK, then if we think about it, that means we're gonna set F of X, natural log of 3 X2 + 5, to be equal to 0. OK. If that's the case, then this implies that 3 X2 + 5 is gonna be equal to 1. But since 3 X 2, OK. Plus 5 is always greater than 1, OK? Because notice here our X quad term is always gonna make it positive and it's always gonna be something that's greater than 5, then that means there are no X intercepts for our graph. Now finally, we have to think about the functions and behavior. So that's point number 6. What's the function doing at its ends? Well, if we think about it, as X approaches 0, OK, then, uh, as X approaches infinity, as X approaches infinity, then F of X is going to be approaching the natural log of 3 X2, OK, which is equal to the natural log of 3 plus the natural log of X2, and we know that that will also be approaching infinity. On the other hand, As X approaches negative infinity, then F of X, because our function is even, will still also be approaching infinity. OK? So that's what will be happening at the ends of our function. Now, we, we've spoken about quite a few things. So let's sum up those properties that we'll need to plot our graph. First, we know that our function grows infinitely as X approaches negative or positive infinity. We know that there are no X intercepts and the Y intercept, the local minimum is 0, natural log of 5. We know that it's increasing. Our function is increasing when X is greater than 0 and decreasing when X is less than 0. It has inflection points at the square root of plus or minus the square root of 5 divided by 3. The function is symmetric above the y axis, and we also know it's concavity, OK. So with all of this in mind, we can plot our points. Now, if we go to our graph here, OK, then first, let's plot our critical point. To 00 LN 5, which is just gonna be a little bit above above 1, OK? It's gonna be somewhere here. It's gonna be 0 LN 5. OK. Next, let's plot our inflection points plus our minus the square of 5 divided by 3. So on the left hand side, it's gonna be a little bit over here. OK, just a little above or a little bit past 1. And on our, sorry, that was on our right hand side. On our left hand side, it's just gonna be a little bit outside of -1. OK, so here. I know that we have these 3 points. OK. Let me just put the values of our points here. It's negative route 5 divided by 3, Ellington. OK. Now we can plot and we know that it's gonna look something like this. So this is a graph of F of X. Thanks for watching everyone. I hope this video helped.

Graphing functions Use the guidelines of this section to make a complete graph of f.

f(x) = ln (x² + 1)

Key Concepts

Natural Logarithm

Graphing Techniques

Behavior of Functions

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