# Calculus Problems Part-Two

1. Mar 30, 2008

### RedBarchetta

Once again, thanks to the physics forum, my previously answered questions helped me get an A on my homework. Thanks! Once again, any help or direction is appreciated.

F(x) = (x^(4/5))*(x - 8)^2
So I tried to take the derivative but came out with a sloppy mess.
First I expanded the last term.
F'(x)=(x^4/5)*(x^2-16x+64)
Applied the product rule.
F'(x)=(4/5)x^(9/5))-(64/5)x^(4/5)+2x^(9/5)-16x^(4/5)
Simplified.
F'(x)=(256/5)x^(-1/5)+(14/5)x^(9/5)-(16/5)x^(4/5)
That's where I get stuck.
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2.Consider the following figure.

Use the graph of f to estimate the values of c to the nearest 0.1 that satisfy the conclusion of the Mean Value Theorem for the interval [1,7]. (4 answers "c=___")

I'm lost on this one.
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3. Find the number c that satisfies the conclusion of the Mean Value Theorem on interval [1, 8].

With this one I started by taking the derivative of the function and ended up with 4x/16.
I set this equal to zero and had x=0. I tried this but it was wrong.
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4. If f(4) = 15 and f '(x) ≥ 2 for 4 ≤ x ≤ 8, how small can f(8) possibly be?

No clue on the theory on this one.
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5. Suppose the derivative of a function f is given below. On what interval is f increasing? (Enter the interval that contains smaller numbers first. If you need to use - or , enter -INFINITY or INFINITY.)
f '(x) = (x + 1)^2(x - 5)^3(x - 6)^4

I typed it into the calculator, I saw the it was going to infinity, I need to take the derivative to find the critical point?
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6. Consider the equation below. (Give your answers correct to two decimal places.)
f(x) = 7sin(x) + 7cos(x)
0 ≤ x ≤ 2(pie)

How do you take the derivative of that? Also, how do you work with pie? It's so painstakingly time-taking for me. I've had to convert everything. This is how far I got:
f'(x)=7(sin(x)+cos(x))
f'(x)=7(cos(x)-sin(x))
f'(x)=0 when 7(cos(x)-sin(x))=0

(cos(x)-sin(x))=0
cos(x)=sin(x)

then...?

(b) Find the local minimum and maximum values of f.
I'm typing in 1.41 and -1.41 but it says its wrong. Also, when your loo
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7. Find the number c that satisfies the conclusion of Rolle's Theorem.
f(x) = x^3- x^2-2x+6 [0, 2]
I got 1.21 but it says its wrong. Argh.
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8. Find the number c that satisfies the conclusion of Rolle's Theorem.
f(x) = cos(2x) on interval [pie/8,7*pie/8]
So, derivative? Chain rule.
f'(x)=-2sin(2x)
f'(x)=0 when f'(x)=-2sin(2x)=0
-2sin(2x)=0
sin(2x)=0
Then...?

Thanks!

Last edited: Mar 30, 2008
2. Mar 30, 2008

### HallsofIvy

Staff Emeritus
It's hard to tell where you went wrong since your parentheses don't match up. It is simpler to go ahead and multiply the x4/5 into the quadratic first rather than use the product rule: F(x)= x14/5- 16x9/5+ 64 so that F'(x)= (14/5)x9/5- (144/5)x4/5+ (256/5)x-1/5. That's almost what you have- perhaps that "14/5" is a typo for "144/5"?
Now you are looking for critical points- points where the derivative either does not exist or is 0. For what x does that erivative not exist? For what x is it 0?

Do you know what the conclusion of the mean value theorem is? That would be a good place to start! What are f(1) and f(7)? What is (f(7)- f(1))/(7- 1)? Why am I asking those questions?

Why did you set it equal to 0? Again, what does the mean value theorem say?
(The derivative of x/(x+4) certainly is not 4x/16.)

The "theory" is that the derivative is the slope of the tangent line and so tells you how fast a function increases. Suppose the derivative were exactly 2 for all x- that is, that it is the straight line with slope 2 passing through (4, 15). What would y be when x= 8?
Do you see that any curve with slope greater than 2 would be above that line?

?? You are given the derivative!! What is true about the derivative of an increasing function? For what values of x is that true? (For God's sake don't multiply this- you want it factored!)

First read the problem carefully. I see no "question" in what you give here. You say "Consider the equation" and then "Give you answer ..." with out any question. What were asked to do? Find the critical points? If so, then your working is correct. You just need to solve the equation cos(x)= sin(x). Do you know any x for which sine and cosine are equal? (There are 2 of them between 0 and 2pi.) As for working with "pie", you slice it and eat it. For "pi" or "$\pi$" that's just a number. You work with it just as you do with 1 or 2. And, of course, you know sine and cosine of $\pi$!

(b)? Was there no (a)? Was (a) to find the critical points of f above? Where did you get "1.41"? I thought you said you weren't able to solve sin(x)= cos(x)! Did you forget the "7" multiplying sin(x)+ cos(x)?

No, 1.21 is not at all correct. Unfortunately, you don't say how you got that answer so I can't say what you did wrong. What is the conclusion of Rolle's theorem?

For what values of 2x is sine equal to 0 (if x is between pi/8 and 7pi/8, then ex is between pi/4 and 7pi/4)? Surely you know something about trigonometry.

3. Mar 30, 2008

### RedBarchetta

I've figured out every problem now minus one. Thanks for the help Ivy!

Find the critical numbers of the function on the interval.
Find:
theta=_____(Smaller value)
theta=_____(larger value)
0<or=to(theta)is<2pi

f(theta)=2cos(theta)+sin(x)^2

I started by taking the derivative of the function.
f'(theta)=7cos(theta)-7sin(theta)
f'(theta)=0 when 7cos(theta)-7sin(theta)=0

7cos(theta)-7sin(theta)=0
7cos(theta)=7sin(theta)
cos(theta)=sin(theta)

Using the unit circle I found that, theta=pi/4 and 5pi/4.

The homework checker said this was wrong. What am I missing here?