3 free response practice AP problems

[Shay10825]

Hi everyone. I need some help with these 3 problems. Any help would be appreciated.

4. Let f be the function defined by f(x)= ln(2+ sin x) for pi <= x <= 2pi

(a) Find the absolute maximum value and the absolute minimum value of f. Show your analysis that leads to your conclusion.

(b) Find the x-coordinate of each point on the graph of f. Justify your answer.

My work:

f'(x) = (cos x) / (sin x +2)
0= (cos x) / (sin x +2)
0= cos x
x= pi/2

f''(x) = [ (sin x +2)(-sin x ) - (cos x)(cos x) ] / [ (sin x +2)^2 ]
0= (sin x +2)(-sinx) - (cos x)(cos x)

Now here is my problem. How do I solve that??

5. The figure above shows the graph of f', the derivative of a function f. The domain of f is the set of all such that 0<x<2

(a) Write an expression for f'(x) in terms of x
(b) Given that f(1)=, write an expression for f(x) in terms of x
(c) In the xy-plane provided below, sketch the graph of y=f(x)

My Work:

a. for (1,1) and (2,0)

y-0=-1(x-2)
y=-x+2

f'(x) = {x 0<= x <= 1
{-x+2 1<= x <=2

b. 0= .5(1^2) +c
-.5 = c

0= 2(1) -.5(1^2) +c
0=2-.5 +c
-1.5=c

f(x)= {.5x^2 + -.5 0<= x <=1
...{2x - .5x^2 -1.5 1<= x <2


6. Let P(t) represent the number of wolves in a population at time t years, when t>= 0. The population P(t) is increasing as a rate directly proportional to 800-P(y), where the constant of proportionality is k.

(a) If P(0)= 500, find P(t) in terms of t and k
(b) If P(2)=700, find k
(c) Find lim P(t)
...t->x

i have no clue what to do for this one. since it says "rate" it must use related rates but how would you do it??

~Thanks[/color]

 

[whozum]

#4, to find the absolute maximum and minimum of a graph, only the first derivative is required. The second derivative is used to find curvature.

Solving the first derivative will give you the points where a critical point occurs. From there, you can look at the graph to see if it is a maximum or minimum. Don't forget to evaluate the endpoints of the interval.

#5. I don't see a question

#6. Thats just straight up confusing.

 

[Shay10825]

I'm sorry about that.

Here is the graph for #5: http://img176.exs.cx/my.php?loc=img176&image=apgraph1vf.png[/color]

 

[whozum]

On #5. there's still a lot of blanks in the question, you might want to re write it but
a) the second part of the composite function is 1-x, not 2-x. You are starting from y = 1, not 2.

b) i dontk now what f(1) is, but it looks like you have the right idea.

c) Can't see

6. That question doesn't make sense, did you write it correctly?

 

[Shay10825]

For number 5 part b they did not give me a picture of the "xy-plane provided below"

For number 4 would I have to solve the equation 0= (sin x +2)(-sinx) - (cos x)(cos x) to get the points of inflection?

Yeah I wrote number 6 correctly[/color]

 

[whozum]

Solving f''(x) will give you the concavic tendencies of the curve.

#6 doesn't make sense to me.

 

[Shay10825]

Yeah I don't understand number 6 either.

How would I find the points of inflection for number 4?[/color]

 

[whozum]

4. Let f be the function defined by f(x)= ln(2+ sin x) for pi <= x <= 2pi

(a) Find the absolute maximum value and the absolute minimum value of f. Show your analysis that leads to your conclusion.

f(x)= ln(2+ sin x) and condition \pi &lt;= x &lt;= 2\pi

f&#039;(x) = \frac{cosx}{sinx+2}

\frac{cosx}{sinx+2} = 0

cosx = 0, x = \frac{3\pi}{2}

There is a critical point at x = \frac{3\pi}{2}

You could look at a graph of the function to see if it is a minimum or maximum, but if you can't do that, then take a point to the left and right of it and evaluate the function there.
f&#039;(x) = \frac{cosx}{sinx+2}

Notice f'(x) for \pi &lt; x &lt; 3\pi/2 is always negative.
Notice f'(x) for 3\pi/2 &lt; x &lt; 2pi is always positive.

The curve is decreasing up to x = 3pi/2, and increasing from 3pi/2 to 2pi. You can conclude that x = 3pi/2 is the absolute minimum of the graph.

Evaluating critical points & endpoints:
f(3\pi/2) = ln(2+sin(3\pi/2)) = ln(2-1) = ln(1) = 0

f(\pi) = ln(2+sin(pi)) = ln(2)

f(2\pi) = ln(2+sin(2\pi)) = ln(2)

The local minimum is at x = 3pi/2, and the value is 0. The local maximums are at x = pi, 2pi and have value ln(2).

 

[Shay10825]

But would those points be the same thing as the points of inflection?

 

[whozum]

It's not asking you for points of inflection. For those you would need to solve f''(x) = 0.

 

[Shay10825]

How would you find part b of number 4?

 

[whozum]

Thats a starngely asked question but:

Part a) min f = 0, max f = ln2
Part b) min x = 3pi/2, max x = pi,2pi

 

[Shay10825]

would the answers to part a be : min x = 3pi/2, max x = pi,2pi because part a is asking you for the min and max?

 

[Shay10825]

Oh my gosh I'm so sorry. For number 4 part b is suppose to read :
(b) Find the x-coordinate of each inflection point on the graph of f. Justify your answer.

 

[whozum]

For part b solve f''(x) = 0. The solutions to that equation are the points of inflection.

 

[Shay10825]

f''(x) = [ (sin x +2)(-sin x ) - (cos x)(cos x) ] / [ (sin x +2)^2 ]
0= (sin x +2)(-sinx) - (cos x)(cos x)

but how would i solve that?

 

[whozum]

\frac{-(sinx+2)sinx-cos^2x}{(sinx+2)^2} = 0

\frac{-sin^2x-2sinx-cos^2x}{(sinx+2)^2} = 0

\frac{-(sin^2x+cos^2x+2sinx)}{(sinx+2)^2} = 0

sin^2x+cos^2x = 1

\frac{-2sinx-1}{(sinx+2)^2} = 0

You should be able to go from there.

 

[Shay10825]

ok I got -pi/6 as my answer. what would the answer be in order to fit in the domain given?

 

[whozum]

The sin function is periodical by 2pi

sin(x) = sin(2n\pi+x)

There is an integer value n where the function will fall in the domain.

 

[Shay10825]

ok I understand now. Thank you so much for helping me :)!

I have 1 more question.

I think I got number 6 now. I got p(t) = -300e^(-kt) + 800
and part c says find:

lim P(t)
t->x

how would I do this?

 

[whozum]

I don't get 6 because its not clear in defining its variables.
But the limit of p(t) = -300e^(-kt) + 800 as t->x = p(x) = -300e^(-kx) + 800

 

[Shay10825]

How do you know that? Is it a rule?

 

[whozum]

limit f(x) = f(a) for a continuous function.
x->a

 

[Shay10825]

Thank you soooo much for your help! :-)

 

[HallsofIvy]

"For part b solve f''(x) = 0. The solutions to that equation are the points of inflection."

Which was, after all, the problem he had originally!

"f''(x) = [ (sin x +2)(-sin x ) - (cos x)(cos x) ] / [ (sin x +2)^2 ]
0= (sin x +2)(-sinx) - (cos x)(cos x)"

= -sin²(x)- 2sin(x)- cos²(x)
= -1- 2sin(x)
sin(x)= -1/2. Can you solve that?

 

[whozum]

In the original problem it didnt say to find the points of inflection, but the critical points. Only later on did he mention that the poitns of inflection were needed, I thought he was just confusing points of inflection and critical points.

 

[leon1127]

Hi everyone. I need some help with these 3 problems. Any help would be appreciated.

4. Let f be the function defined by f(x)= ln(2+ sin x) for pi <= x <= 2pi

(a) Find the absolute maximum value and the absolute minimum value of f. Show your analysis that leads to your conclusion.

(b) Find the x-coordinate of each point on the graph of f. Justify your answer.

My work:

f'(x) = (cos x) / (sin x +2)
0= (cos x) / (sin x +2)
0= cos x
x= pi/2

f''(x) = [ (sin x +2)(-sin x ) - (cos x)(cos x) ] / [ (sin x +2)^2 ]
0= (sin x +2)(-sinx) - (cos x)(cos x)

Now here is my problem. How do I solve that??

5. The figure above shows the graph of f', the derivative of a function f. The domain of f is the set of all such that 0<x<2

(a) Write an expression for f'(x) in terms of x
(b) Given that f(1)=, write an expression for f(x) in terms of x
(c) In the xy-plane provided below, sketch the graph of y=f(x)

My Work:

a. for (1,1) and (2,0)

y-0=-1(x-2)
y=-x+2

f'(x) = {x 0<= x <= 1
{-x+2 1<= x <=2

b. 0= .5(1^2) +c
-.5 = c

0= 2(1) -.5(1^2) +c
0=2-.5 +c
-1.5=c

f(x)= {.5x^2 + -.5 0<= x <=1
...{2x - .5x^2 -1.5 1<= x <2


6. Let P(t) represent the number of wolves in a population at time t years, when t>= 0. The population P(t) is increasing as a rate directly proportional to 800-P(y), where the constant of proportionality is k.

(a) If P(0)= 500, find P(t) in terms of t and k
(b) If P(2)=700, find k
(c) Find lim P(t)
...t->x

i have no clue what to do for this one. since it says "rate" it must use related rates but how would you do it??

~Thanks[/color]

on the first question, u have to check the end point.