24–34. Curve sketching Use the guidelines given in Section 4.4...

Question

24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.
ƒ(x) = 10x² / (x² + 3)

ƒ(x) = 10x² / (x² + 3)

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Draw the graph of the function F of X is equal to 5 x 2 divided by x 2 + 5, using the information given below. F of X is equal to 50 x divided by x 2 + 5 all the power of 2. F of X is equal to 50 multiplied by -3 x 2 + 5, all divided by X2 + 5, all the power of 3. Fantastic. So it appears for this particular problem, we're asked to graph the function F of X using the information that's provided to us, and the information that's provided to us is we're told what the 1st and 2nd derivative of this function of F of X is. So using our 1st and 2nd derivative derivatives, I should say for F of X, we're asked to figure out how to graph F of X. So with that said, now that we know what we're ultimately trying to solve for. We need to first determine the following information to help us plot this graph. So the first piece of information that we need to determine in order to help us properly graph our function of F of X is we need to determine the domain, which I'll call this D for domain. So as I should note, the function F of X is equal to 5 x 2 divided by x 2 + 5. is defined for all real values of X because the denominator X2 + 5 is always positive and it can never be zero since, as we should note that x 2. Once again, we need to note that X to the power 2 is greater than or equal to 0 for all real numbers. So therefore, the domain is all real numbers, so you can use your fancy little R symbol to say for all real numbers. Fantastic. So our second step now, now that we know that the domain is all real numbers, we now need to determine the symmetry, which I'll call S. So as we should recall, to check the symmetry, we need to determine if the function is symmetric about the Y axis, which will make it an even function, or if it's symmetric about the origin, which would make it an odd function, or if It has no symmetry at all. So, as we should recall, it will have even symmetry when F of negative X is gonna be equal to F of X, and it's gonna be odd when F of negative X is equal to negative F of X. Fantastic. So now we need to take and plug in a value of negative X in for x for our original functions. So we need to take F of negative X is equal to. So we take 5 multiplied by negative X to the power of 2, all divided by negative X2 + 5, we will find that F of negative X is going to be equal to. 5 x to the power of 2 divided by X2 + 5, which is going to be equal to F of X. Fantastic, moving right along here. So therefore, F of X is even and we'll have symmetry about the Y axis. So let's make a note of that. So it's gonna be even, so an E for even, and it's going to have symmetry about the Y axis. Fantastic. So now our next step now, our third step is we need to determine the critical points. So as we should recall a note, in order to find the critical points, we need to set F of X equal to 0. So we're told that F of X is equal to 50 x divided by X2 + 5 all to the power 2, and we need to set this equal to 0. So, we need to note that since the numerator is 50 X, it will equal 0 when X equals 0, so therefore, the only critical point is going to be at X equals 0 as a result. Now, we need to solve for our 4th step, which is the inflection points, which I'll call IP. And in order to find the inflection points, as we should recall, we need to set F of X equal to 0. So, as we could see, When we take our function and we set F of X equal to 0, we will obtain the following, so we'll get 50. Multiplied by 3 x 2 + 5, all divided by X2 + 5, all to the power of 3. is equal to 0, and what we're trying to do now is we're trying to rearrange this equation to isolate and solve for X all by itself, which I'm going to skip several steps by using algebra and rearranging it just to isolate and solve for X by itself, we will find that X is equal to plus or minus the square root of 15 divided by 3. So now we need to find the Y coordinates of all the inflection points. So we need you to take F of the square root of 15 divided by 3 is equal to, so that's when you plug in a value of square root of 15 divided by 3 in for X in our to our original F of X function. And when we do, we will find that it's equal to 5 divided by 4, and once again you're taking. Our value of X and you're plugging it into our original function for all values of X, and when you solve, you should get 5 divided by 4. And similarly, if we take F of negative square root of 15 divided by 3. And we take that value of X and we plug it into our original function, we will find that it's also equal to 5 divided by 4. Fantastic. So now, Our next step is we need to find the increasing and decreasing intervals. So we need to know that the function is increasing when F Of X is greater than 0, so that means it's increasing, and it will be decreasing when F of X is less than 0. It will be decreasing. So, from the first derivative F of X is equal to 50 x divided by X2 + 5 all to the 2, we will see that the function is increasing for X is greater than 0. And decreasing for X is less than 0. So let's make a note of that. So, X is greater than 0, it's going to be increasing, and when X is less than 0, it's going to be decreasing. So now, our next step. is we need to determine If it's concave up or concave down, so it's gonna be concave up or concave down. So the function will be concave up where F of X is going to be, so F of X is gonna be greater than 0, and it's gonna be Poca, so let's make a note of that. So it's gonna be. Let's keep this simple. So it's gonna be concave up, so let's make a note. It's gonna be concave up. When F prime of X. So F of X is going to be greater than 0, and it's going to be concave down where F of X is less than 0. Fantastic. So with that said, so when we analyze the second derivative, it will help us determine the concavity intervals. So we need to note that it's going to be concave up. When parentheses negative, square root of 15 divided by 3, 15 divided by 3 in parentheses, and it's going to be concave down when it's negative infinity negative, square root of 15 divided by 3 in parentheses is in union with square root of 15 divided by 3, infinity. Fantastic. So then our next step is we need to find the extreme values and the inflection points. So that's our 7th step, so let's make a note since we're running out of room here. So our 7th step is we're trying to find the extreme values. So the extreme values and inflection points, so we need to note that there's an extreme value at X equals 0, which you can classify by evaluating F of X at that point, as we should recall, and checking the sign of F of X. And if F of X changes the sign at X equals 0, then there will be an inflection point present. And our 8th step is we need to look at the asymptotes and and behavior. So the function has no vertical asymptotes because the denominator X2 + 5 will never equal 0 or never B 0. So horizontal asymptotes such as X approaches. Infinity and F of X approaches 5 x 2 divided by X2 equals 5. So the function will have a horizontal asymptote at Y equals 5. So once again, it has a horizontal asymptote at Y equals 5. Fantastic. So the intercepts, we need to figure out our intercepts. So to find the Y intercept we need to set X equal to 0. So when we plug in a value of F equals 0 or we plug in a value 0 in for x, we will find that it's going to be simply equal to 0. So the Y intercept is going to be at parentheses 0.0, which once again, as we should recall, that's the origin point. And for the X intercept, we need to set F of X equal to 0. So we take our original function and set it equal to 0 and rearrange it to isolate and solve for X. And when we do, we will find that X is simply equal to 0. So that means the X intercept is also at parentheses 0.0, which is the same as our Y intercept. So now that we've found all of our properties that we obtained and we have just obtained together, we can now officially graph our final answer. So looking at our blank graph, which was provided to us by the problem itself, using the information that we determined together, we should produce the following graph. So, as you could see, we have our horizontal asymptote at, once again, we're told that our horizontal asymptote is at Y equals 5, which we have represented by a dotted yellow line. And of course, our curve, which is represented in blue, gets super, super close to this horizontal asymptope, but it'll never touch or cross this horizontal asymptote. So as you can see, it starts in the negative quadrant to the left, and the curve follows the horizontal asyote very closely, and then it dips down where it touches the origin point, and then it starts to go up, so it forms like a parabolic shape. And then it goes up and then it starts to go closer and closer and closer again to the horizontal asymptote. So it almost looks like the like a bird in flight, like a seagull up high in the sky is what the shape of our curve looks like. And then as you can see on our graph, we have our two inflection points, and then we have where our X and Y intercepts are, which is at 0.0, and they're all represented by yellow dots. And once again, we have our inflection point, which is negative the square root of 15 divided by 3.5 divided by 4 in parentheses, and then we have an inflection point at positive, square root of 15 divided by 3.5 divided by 4 in parentheses. And that's it. We've officially graphed this problem, which we have solved for our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.
ƒ(x) = 10x² / (x² + 3)

Key Concepts

Function Behavior

Critical Points and Derivatives

Graphing Techniques

Watch next

24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.ƒ(x) = 10x² / (x² + 3)

Key Concepts

Function Behavior

Critical Points and Derivatives

Graphing Techniques

Watch next

24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.
ƒ(x) = 10x² / (x² + 3)