An expression that occurs in calculus is given...

[sweiss]

Not sure exactly what this should look like once it is solved. Here is the question as found on paper:


An expression that occurs in calculus is given. Write the expression as a single quotient in which only positive exponents and/or radicals appear.

(x^2-1)^1/2 * 3/2(2x+5)^1/2 * 2 + (2x+5)^3/2 * 1/2(x^2-1)^-1/2 * 2x

I understand it is somewhat of a factoring problem. Here is the solution I am ending up with...

[(2x+5)^1/2(x^2+5x-3)] / (x^2-1)^1/2

Somehow this doesn't seem like the right solution! Any help is appreciated.

Thanks,

S. Weiss

 

[Deleted member 4993]

Not sure exactly what this should look like once it is solved. Here is the question as found on paper:


An expression that occurs in calculus is given. Write the expression as a single quotient in which only positive exponents and/or radicals appear.

(x^2-1)^1/2 * 3/2(2x+5)^1/2 * 2 + (2x+5)^3/2 * 1/2(x^2-1)^-1/2 * 2x

I understand it is somewhat of a factoring problem. Here is the solution I am ending up with...

[(2x+5)^1/2(x^2+5x-3)] / (x^2-1)^1/2

Somehow this doesn't seem like the right solution! Any help is appreciated.

Thanks,

S. Weiss

Choose an arbitrary value for x (other than x=+ or -1) - say x=2 (use a spreadsheet program like excel) - and evaluate the given function and the one you derived. If both of those give you the same answer, you are most probably correct.

 

[sweiss]

Thanks for the reply!

I tried that and am getting vastly different answers for each. Now I am just not sure how to begin on completing the problem anew...

 

[Dr.Peterson]

(x^2-1)^1/2 * 3/2(2x+5)^1/2 * 2 + (2x+5)^3/2 * 1/2(x^2-1)^-1/2 * 2x

I understand it is somewhat of a factoring problem. Here is the solution I am ending up with...

[(2x+5)^1/2(x^2+5x-3)] / (x^2-1)^1/2

Somehow this doesn't seem like the right solution! Any help is appreciated.

I find that only the term in bold is wrong. Check your work step by step to correct it. I think you probably did the right things here, and just missed a sign or something.

If you can't find the error, you can show us an image of your work itself.

I should add, you need to use parentheses more carefully. I would write this for clarity:

(x^2-1)^(1/2) * (3/2)(2x+5)^(1/2) * 2 + (2x+5)^(3/2) * (1/2)(x^2-1)^(-1/2) * 2x​

 

[sweiss]

I find that only the term in bold is wrong. Check your work step by step to correct it. I think you probably did the right things here, and just missed a sign or something.

If you can't find the error, you can show us an image of your work itself.

I should add, you need to use parentheses more carefully. I would write this for clarity:

(x^2-1)^(1/2) * (3/2)(2x+5)^(1/2) * 2 + (2x+5)^(3/2) * (1/2)(x^2-1)^(-1/2) * 2x​

I think I did find an error where you bolded, it should be a 5x^2 there. However, even with this corrected, when I choose an arbitrary number for x, like 2, as the above forum moderator suggested, I still get vastly different answers. I will go ahead and post a picture of my work to see if anything else pops out to you. Thanks so much!

 

[Dr.Peterson]

Your work is identical to mine.

I imagine you are evaluating one or both of the expressions incorrectly, which is easy to do. Can you show us the number you chose, and your results, so we can check those? I do get the same value for both expressions for x=2, doing it on paper and exactly. A calculator may make it harder!

 

[sweiss]

Your work is identical to mine.

I imagine you are evaluating one or both of the expressions incorrectly, which is easy to do. Can you show us the number you chose, and your results, so we can check those? I do get the same value for both expressions for x=2, doing it on paper and exactly. A calculator may make it harder!

Well that first statement is good to hear! I'm sure that I am just messing something up with the rational exponents or something. I went ahead and did it on paper using x=2 and am still getting different answers. I'll attach that here.

 

[Dr.Peterson]

Check your second line. You dropped the 1/2 from the line above.

If you're wondering why I'm so good at spotting arithmetic mistakes, it's because I've had lots of practice ... on my own work.

 

[sweiss]

Check your second line. You dropped the 1/2 from the line above.

If you're wondering why I'm so good at spotting arithmetic mistakes, it's because I've had lots of practice ... on my own work.

I think I might have multiplied the 1/2 by the 2x at then end of that line in the first row which reduced to the 2 that is at the front of those parenthesis in the second row. Not sure if that was the right thing to do though.

 

[Dr.Peterson]

I think I might have multiplied the 1/2 by the 2x at then end of that line in the first row which reduced to the 2 that is at the front of those parenthesis in the second row. Not sure if that was the right thing to do though.

Well, 1/2 times 2 is 1, but you kept the 2! No, I see, the 2 is your x.

One thing I would definitely have done differently is to move things around, and substitute the value, in separate steps. That makes it easier to see what you did.

Now that I am looking past the mistake I thought I saw, I see a horrendous mistake on the fourth line: What is 9^(3/2)? It is not 9^(2/3)! There are no cube roots in this problem.

 

[sweiss]

Well, 1/2 times 2 is 1, but you kept the 2! No, I see, the 2 is your x.

One thing I would definitely have done differently is to move things around, and substitute the value, in separate steps. That makes it easier to see what you did.

Now that I am looking past the mistake I thought I saw, I see a horrendous mistake on the fourth line: What is 9^(3/2)? It is not 9^(2/3)! There are no cube roots in this problem.

That was it! 9^(3/2)=27! When I corrected that, both of my answers come out identical. Thanks so much!

 

[JeffM]

Ugly things like

[MATH]g(x) = (x^2 - 1)^{1/2} * \dfrac{3}{2} * (2x + 5)^{1/2} * 2 + (2x + 5)^{3/2} * \dfrac{1}{2} * (x^2 - 1)^{-1/2} * 2x.[/MATH]
come up all the time in calculus. The algebra in calculus may become messy, and it is easy to make mistakes. (My algebra really improved when I studied calculus.) But they teach an algebraic trick in calculus that REALLY should be taught in algebra. They call it a u-substitution although you can use any letter.

[MATH]\text {Let } u = (x^2 - 1) \text { and } v = (2x + 5) \implies \\ g(x) = u^{1/2} * \dfrac{3}{\cancel 2} * v^{1/2} * \cancel 2 + v^{3/2} * \dfrac{1}{\cancel 2} u^{-1/2} * \cancel 2x \implies \\ g(x) = u^{1/2}v^{1/2}(3 + v^1u^{-1}x) = u^{1/2}v^{1/2} \left (3 + \dfrac{vx}{u} \right ) \implies \\ g(x) = u^{1/2}v^{1/2} * \dfrac{3u + vx}{u} = \sqrt{uv} * \dfrac{3u + vx}{u}.[/MATH]You are unlikely to screw up the polynomials in x in the above. You are just worried about cancelling constants and getting the exponents right.

Now check.

[MATH](uv)^{1/2} * \dfrac{3u + vx}{u} = u^{1/2}v^{1/2} * \dfrac{3u + vx}{u} = u^{-1/2}v^{1/2}(3u + vx) =\\ 3u^{1/2}v^{1/2} + u^{-1/2}v^{3/2}x = 2 * \dfrac{3}{2} * u^{1/2}v^{1/2} + 2 * \dfrac{1}{2} * u^{-1/2}v^{3/2}x.\ \checkmark[/MATH]Now get replace u and v and simplify.

[MATH]g(x) = \sqrt{(x^2 - 1)(2x + 5)} * \dfrac{3(x^2 - 1) + (2x + 5)x}{x^2 - 1} \implies \\ g(x) = \sqrt{2x^3 + 5x^2 - 2x - 5} * \dfrac{5x^2 + 5x - 3}{x^2 - 1}.[/MATH]Now check numerically.

[MATH]g(2) = (2^2 - 1)^{1/2} * \dfrac{3}{2} * (2(2) + 5)^{1/2} * 2 + (2(2) + 5)^{3/2} * \dfrac{1}{2} * (2^2 - 1)^{-1/2} * 2(2) = \\ (4 - 1)^{1/2} * \dfrac{3}{\cancel 2} * (4 + 5)^{1/2} * \cancel 2 + (4 + 5)^{3/2} * \dfrac{1}{\cancel 2} * (4 - 1)^{-1/2} * \cancel 2(2) =\\ \sqrt{3} * 3 * \sqrt{9} + (\sqrt{9})^3 * \dfrac{2}{\sqrt{3}} =\\ 9 \sqrt{3} + \dfrac{2 * 27}{\sqrt{3}} = 9 \sqrt{3} + \dfrac{2 * 27\sqrt{3}}{3} = 27\sqrt{3}.[/MATH][MATH]g(2) = \sqrt{2(2^3) + 5(2^2) - 2(2) - 5} * \dfrac{5(2^2) + 5(2) - 3}{2^2 - 1} \implies \\ g(2) = \sqrt{16 + 20 - 4 - 5} * \dfrac{20 + 10 - 3}{3} = \sqrt{27} * \dfrac{27}{3} =\\ 9\sqrt{9 * 3} = 27\sqrt{3}. \ \checkmark[/MATH]

 

[Steven G]

I just want to make a comment. You should generally try using x=1 but in this problem that makes one of the terms undefined so you can't use x=0.

 

[sweiss]

Ugly things like

[MATH]g(x) = (x^2 - 1)^{1/2} * \dfrac{3}{2} * (2x + 5)^{1/2} * 2 + (2x + 5)^{3/2} * \dfrac{1}{2} * (x^2 - 1)^{-1/2} * 2x.[/MATH]
come up all the time in calculus. The algebra in calculus may become messy, and it is easy to make mistakes. (My algebra really improved when I studied calculus.) But they teach an algebraic trick in calculus that REALLY should be taught in algebra. They call it a u-substitution although you can use any letter.

[MATH]\text {Let } u = (x^2 - 1) \text { and } v = (2x + 5) \implies \\ g(x) = u^{1/2} * \dfrac{3}{\cancel 2} * v^{1/2} * \cancel 2 + v^{3/2} * \dfrac{1}{\cancel 2} u^{-1/2} * \cancel 2x \implies \\ g(x) = u^{1/2}v^{1/2}(3 + v^1u^{-1}x) = u^{1/2}v^{1/2} \left (3 + \dfrac{vx}{u} \right ) \implies \\ g(x) = u^{1/2}v^{1/2} * \dfrac{3u + vx}{u} = \sqrt{uv} * \dfrac{3u + vx}{u}.[/MATH]You are unlikely to screw up the polynomials in x in the above. You are just worried about cancelling constants and getting the exponents right.

Now check.

[MATH](uv)^{1/2} * \dfrac{3u + vx}{u} = u^{1/2}v^{1/2} * \dfrac{3u + vx}{u} = u^{-1/2}v^{1/2}(3u + vx) =\\ 3u^{1/2}v^{1/2} + u^{-1/2}v^{3/2}x = 2 * \dfrac{3}{2} * u^{1/2}v^{1/2} + 2 * \dfrac{1}{2} * u^{-1/2}v^{3/2}x.\ \checkmark[/MATH]Now get replace u and v and simplify.

[MATH]g(x) = \sqrt{(x^2 - 1)(2x + 5)} * \dfrac{3(x^2 - 1) + (2x + 5)x}{x^2 - 1} \implies \\ g(x) = \sqrt{2x^3 + 5x^2 - 2x - 5} * \dfrac{5x^2 + 5x - 3}{x^2 - 1}.[/MATH]Now check numerically.

[MATH]g(2) = (2^2 - 1)^{1/2} * \dfrac{3}{2} * (2(2) + 5)^{1/2} * 2 + (2(2) + 5)^{3/2} * \dfrac{1}{2} * (2^2 - 1)^{-1/2} * 2(2) = \\ (4 - 1)^{1/2} * \dfrac{3}{\cancel 2} * (4 + 5)^{1/2} * \cancel 2 + (4 + 5)^{3/2} * \dfrac{1}{\cancel 2} * (4 - 1)^{-1/2} * \cancel 2(2) =\\ \sqrt{3} * 3 * \sqrt{9} + (\sqrt{9})^3 * \dfrac{2}{\sqrt{3}} =\\ 9 \sqrt{3} + \dfrac{2 * 27}{\sqrt{3}} = 9 \sqrt{3} + \dfrac{2 * 27\sqrt{3}}{3} = 27\sqrt{3}.[/MATH][MATH]g(2) = \sqrt{2(2^3) + 5(2^2) - 2(2) - 5} * \dfrac{5(2^2) + 5(2) - 3}{2^2 - 1} \implies \\ g(2) = \sqrt{16 + 20 - 4 - 5} * \dfrac{20 + 10 - 3}{3} = \sqrt{27} * \dfrac{27}{3} =\\ 9\sqrt{9 * 3} = 27\sqrt{3}. \ \checkmark[/MATH]

Thanks for taking the time to work through that for me. I am vaguely familiar with the concept of substitution, but definitely didn't think to utilize that within this problem. Definitely will be something helpful to keep in mind for the future. Thanks again!

 

[sweiss]

I just want to make a comment. You should generally try using x=1 but in this problem that makes one of the terms undefined so you can't use x=0.

That makes sense. So in this particular equation, since x=1 would make a denominator 0, x=2 is the next best choice?

 

[Steven G]

That makes sense. So in this particular equation, since x=1 would make a denominator 0, x=2 is the next best choice?

Yes! Again, I would always try x=1 first unless it is not in the domain (it makes the denominator 0 or it makes what is under the square root sign negative)

 

[JeffM]

Thanks for taking the time to work through that for me. I am vaguely familiar with the concept of substitution, but definitely didn't think to utilize that within this problem. Definitely will be something helpful to keep in mind for the future. Thanks again!

Substitution becomes almost automatic in calculus. Actually in calculus, you might well have done the substitution before you ever get to algebra. This particular problem looks as though it is the back end of a calculus problem about finding the maximum or minimum of the function below.

[MATH]f(x) = (x^2 - 1)^{1/2}(2x - 5)^{3/2} = u^{1/2}v^{3/2}.[/MATH]
To find the minima and maxima of that function, you must locate at what values of x they occur. Those values are determined by a related function called the derivative. The derivative of products and powers are found by simple rules that are usually stated in terms of u and v. So the way you memorize the rules makes substitution second nature. But it is a useful trick in algebra as well. Solve

[MATH]2(x + 1)^6 + 8(x + 1)^3 = - 8.[/MATH]
Now you may see the solution, or you may say "polynomial of 6 degrees may have no solution in radicals." If, however, substitution is in your bag of tricks, you go

[MATH]u = (x + 1)^3 \implies 2u^2 + 8u = - 8 \implies u^2 + 4u + 4 = 0 \implies \\ (u + 2)^2 = 0 \implies u = - 2 \implies (x + 1)^3 = - 2 \implies x = - (1 + \sqrt[3]{-2}).[/MATH]Let's check

[MATH]\{- (1 + \sqrt[3]{-2}) + 1\}^3 = (\sqrt[3]{-2})^3 = - 2 \implies \\ \{- (1 + \sqrt[3]{-2}) + 1\}^6 = (\{- (1 + \sqrt[3]{-2}) + 1\}^3)^2 = (-2)^2 = 4.\\ 2 * 4 + 8(-2) = 8 - 16 = - 8.\ \checkmark[/MATH]

 

[Petru21]

Choose an arbitrary value for x (other than x=+ or -1) - say x=2 (use a spreadsheet program like excel) - and evaluate the given function and the one you derived. If both of those give you the same answer, you are most probably correct.

Did it give the same answer?

 

[JeffM]

Did it give the same answer?

Did you try it? What did you get?