Seemingly inconceivable problems.

[Fenn]

Hi there, I'm new to the forum. I'm preparing to take the CLEP test for Calculus I and recently took a practice test, however, there were several problems I had difficulty with, one of which appears to be impossible.

f(x) = x^3 + x, if h is the inverse function of f then h'(2)= ?

As far as I know, there is no inverse of x^3 + x since the derivative, 3x² + 1 cannot be negative.

Help is greatly appreciated.

 

[Fenn]

I'm going to be a bit greedy and post another problem I've been having trouble with in this same post, kill two birds with one stone.

Evaluate the definite integral.

?3
?-3 |x+2| dx.

The answer is 13, but again, I have no idea how to reach this conclusion.

Thanks again!

 

[galactus]

Hi there, I'm new to the forum. I'm preparing to take the CLEP test for Calculus I and recently took a practice test, however, there were several problems I had difficulty with, one of which appears to be impossible.

f(x) = x^3 + x, if h is the inverse function of f then h'(2)= ?

As far as I know, there is no inverse of x^3 + x since the derivative, 3x² + 1 cannot be negative.

Help is greatly appreciated.

\(\displaystyle h'(2)=\frac{1}{f'(h(2))}\)

\(\displaystyle h(2)=1\) because f(1)=2.

\(\displaystyle h'(2)=\frac{1}{f'(1)}\)

But, \(\displaystyle f'(x)=3x^{2}+1\). Thus, \(\displaystyle f'(1)=4\)

and therefore, \(\displaystyle h'(2)=\frac{1}{4}\)

 

[galactus]

I'm going to be a bit greedy and post another problem I've been having trouble with in this same post, kill two birds with one stone.

Evaluate the definite integral.

?3
?-3 |x+2| dx.

The answer is 13, but again, I have no idea how to reach this conclusion.

Thanks again!

You have to integrate the positive and negative.

\(\displaystyle \int_{-3}^{-2}(-x-2)dx+\int_{-2}^{3}(x+2)dx\)

You have to split the integrals at the cusp. Graph it and you can see.

 

[Fenn]

I'm going to be a bit greedy and post another problem I've been having trouble with in this same post, kill two birds with one stone.

Evaluate the definite integral.

?3
?-3 |x+2| dx.

The answer is 13, but again, I have no idea how to reach this conclusion.

Thanks again!

You have to integrate the positive and negative.

\(\displaystyle \int_{-3}^{-2}(-x-2)dx+\int_{-2}^{3}(x+2)dx\)

You have to split the integrals at the cusp. Graph it and you can see.

I'm going to guess that you intended to declare the absolute value of the integral rather than parenthesis.

When I do it this way, I get the answer. Thanks a bunch

Edit*** Oh and also, do I only need to split it when an absolute value is involved or are there other scenarios (concerning Calculus I) that would require me to split the integral between + and - ?

 

[Fenn]

Looking over these problems, there's only one left that truly baffles me, I'd greatly appreciate help on it so that we can put this thread to rest =).

I am given the following chart:

x || f(x) | f '(x) | g(x) | g '(x)
__________________________________
10 || 35  |  15   |    6   |  4
20 ||  8   |   5   |   12   |  10
30 || 24  |  25   |   20  |  10

Sorry for the poorly improvised data table, anyway, if h(x) = f(g(x)) then what does h'(30) = ?

The answer is 50, however, once again I'm lost as to how to reach this conclusion.

 

[galactus]

No, I intended the parentheses on the integration. That is because |x+2| is x+2 or -(x+2)

As for the problem at hand.

Apply the chain rule:

\(\displaystyle h'(x)=f'(g(x))g'(x)\)

\(\displaystyle g(x)=20\), so \(\displaystyle f'(g(x))=f'(20)=5\)

\(\displaystyle g'(x)=10\)

\(\displaystyle 10(5)=50\).

 

[mmm4444bot]

Seemingly inconceivable problems

Do you mean that you don't trust your eyes?

 

[Fenn]

Re:

Seemingly inconceivable problems

Do you mean that you don't trust your eyes?

Nah, not exactly =). Just I haven't come across a single problem like the ones I posted above in my calculus book, except perhaps the initial inverse one. I went to google the inverse of x³+x and found a couple of websites using that exact formula as examples of functions that were impossible to inverse. I'm sure you could understand my confusion as a calc newbie hehe.

Anyway, 12 hours from now I take the real deal, a bit nervous, but i've been doing dozens of CLEP and AP practice tests since I made this thread, if anyone has any last minute recommendations or tips, feel free.

 

[Fenn]

Actually I have one last minute question.

Which of the following integrals have the same value?

?b
?a f(x)dx

?b-a
?0 f(x+a)dx

?b+c
?a+c f(x+c)dx

The answer is the first two are equal, the last one is not.

Yet, I fail to see how the last integral is not equal to the first two, the way I see it is:
(b²/2 + c²/2 +c²/2) - (a²/2 + c²/2 +c²/2). In which case the c's would cancel out, resulting in the integral of b-a. Why am I wrong?