#$&*
course Mth 174
Section 6.11) *6.1.2 Sketch and describe two functions F such that F' = f when f is the line going through (1,0) and (0,1).
A: F(0)=0 The graph starts at (0,0) and is increasing up to x=1. At x=1 the slope is 0 and then the graph starts to decrease towards negative infinity.
F(0)=1 the graph starts at (0,1) and looks just like the first.
2) *6.1.4 Sketch and describe two functions F such that F' = f when f(x) = |x|
A: F(0)=0 The graph is concave down and increasing up to the inflection point at x=1 and is then concave up and increasing.
F(0)=1 The graph starts at (0,1) and is the same as the first.
3) 6.1.6 Estimate f(x) for x = 2,4,6 when f(0) = 50 and f '(0)= 17, f '(2)= 15, f '(4) =10, and f '(6) = 2.
A: Using the fundamental theorem, we find that f(b)=f(a)+the integral of f prime between b and a.
f(2)=f(0)+ (f’(0)+f’(2))/change in x/the interval.=82
f(4)=107
f(6)=119
4) *6.1.8 Let dP/dt be described by the piecewise defined function which follows. When t <= 2, dP/dt = -1. When 2 <= t <= 4, dP/dt = t - 3. When t >= 4, dP/dt = 1. Sketch dP/dt and using P(0) = 2 use dP/dt to find P(t) when t = 1,2,3,4,5.
A: The graph dP/dt. From 0 to 1 the slope of P is -2 so dP/dt is a straight line starting at P=-2 From t=0 to t=1. From t=1 to t=2, the straight line continues to t=2. From t=2 to t=3, the straight line is at -(9/2). From 3 to 4 the graph is at -(9/2). From 4 to 5, the graph is at 2.
Using the fundamental theorem and the fact that P(t) when 2<=t<=4 is (1/2)t^2-3t+2 because c=2 when P(0)=2, we see that P(1)=0, P(2)=-2, P(3)=-(13/2), P(4)=-11, P(5)=-9
5) f(x) is a function defined on the interval 0 <= x <= 7 with the following characteristics:
f '(x) = 1 for all x on the interval (0,2),
f '(x) = -1 for all x on the interval (2,3),
f '(x) = 2 for all x on the interval (3,4),
f '(x) = -2 for all x on the interval (4,6),
f '(x) = 1 for all x on the interval (6,7)
f(3) = 0
What is the integral of f ' (x) over the interval 0 <= x <= 7?
Describe your graph of f(x), indicating where it is increasing and decreasing and where it is concave up, where it is straight and where it is concave down.
What are the following values:
f(2)
f(3)
f(4)
f(6)
f(7)
Is the graph of f(x) continuous?
A: a) The integral of f’(x) between 0 and 7 is f’(7)-f’(0)=1-1=0
b)f(x) is increasing 0<=x<=2, 3<=x<=4, 6<=x<=7
decreasing 2<=x<=3, 4<=x<=6
concave up 2<=x<=4, 4<=x<=7
concave down 0<=x<=3, 3<=x<=6
It’s never straight because the slope is never 0.
c) Using the fundamental therem, we find that
f(2)=1, f(3)=0, f(4)=2, f(6)=1, f(7)=2
d)yes
6) *6.1.10 Let f (x)= x^3 - 3x^2 and F'(x) = f(x). Sketch or graph f(x) (with a graphing calculator if necessary, though you should be able to construct the graph easily) and referring only to the graph of f(x), give the critical points, explain which are local maxima/minima, and which are neither. Also sketch a possible graph of F(x).
A: Critical points at x=0 and 3. L min x=2, L max x=0. Inflection Point x=1.
F(x) is a concave up parabola passing through x=0.
7) *6.1.12 Sketch and describe two functions F with F'(x) = f(x) (Sketch or graph f(x) with a graphing calculator) by only looking at the graph of f(x) = x^3 - 4x^2 + x + 6. In one let F(0) = 0 and the other F(0) = 1. Also identify local minima, maxima, and points of inflection.
A: Local min x=3
Local max x=2
Inflection point x=2.5
8) *6.1.18 Let g' ={-x when 0 <= x <= 10, 4x when 10 <= x <= 15, x when 15 <= x <= 20, and -(1/2)x when x >= 20}. Sketch g' and given g(0) = 50 sketch the graph of g(x) and specify all critical and inflection points of g(x).
A: Inflection point x=10 and x=20
Local min x=13
9) 6.1.22 Use a graph of 4cos(x^2) to determine where an antiderivative, F, of this function reaches its maximum on 0<= x<= 3. If F(1) =4, find this maximum obtained.
A: I don’t know how to find the anti derivative.
10) Given the graph below, which depicts outflow vs. clock time and inflow vs. clock time for a reservoir:
When was the quantity of water greatest and when least? Describe in terms of the behavior of the two curves.
When was the quantity of water increasing fastest, and when most slowly? Describe in terms of the behavior of the two curves.
A: a) July greatest and Jan. 08 least.
between about march and july, the inflow is greater than the outflow showing that the greatest amount of water is accumulated in july and then decreases from there because the outflow is greater than the inflow and continues to be so up to June 08.
b) Little after april (may) fastest, Oct slowest.
the only time the inflow is greater than the out is between march and july and the highest point of the inflow is in may.
Oct is the slowest because the outflow is at its highest point above the inflow."
Self-critique (if necessary):
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Self-critique rating:
#$&*
course Mth 174
Section 6.11) *6.1.2 Sketch and describe two functions F such that F' = f when f is the line going through (1,0) and (0,1).
A: F(0)=0 The graph starts at (0,0) and is increasing up to x=1. At x=1 the slope is 0 and then the graph starts to decrease towards negative infinity.
F(0)=1 the graph starts at (0,1) and looks just like the first.
2) *6.1.4 Sketch and describe two functions F such that F' = f when f(x) = |x|
A: F(0)=0 The graph is concave down and increasing up to the inflection point at x=1 and is then concave up and increasing.
@& Between x = 0 and x = 1 we would have f(x) = | x | = x, which is increasing. So the slope of F(x) would be increasing and F(x) would be concave up on this interval.*@
F(0)=1 The graph starts at (0,1) and is the same as the first.
3) 6.1.6 Estimate f(x) for x = 2,4,6 when f(0) = 50 and f '(0)= 17, f '(2)= 15, f '(4) =10, and f '(6) = 2.
A: Using the fundamental theorem, we find that f(b)=f(a)+the integral of f prime between b and a.
f(2)=f(0)+ (f’(0)+f’(2))/change in x/the interval.=82
f(4)=107
f(6)=119
@& (f ' (0) + f ' (2) ) / 2 = (15 + 17) / 2 = 16. It's hard to see how you got 82 out of that. I need to see more details.*@
4) *6.1.8 Let dP/dt be described by the piecewise defined function which follows. When t <= 2, dP/dt = -1. When 2 <= t <= 4, dP/dt = t - 3. When t >= 4, dP/dt = 1. Sketch dP/dt and using P(0) = 2 use dP/dt to find P(t) when t = 1,2,3,4,5.
A: The graph dP/dt. From 0 to 1 the slope of P is -2 so dP/dt is a straight line starting at P=-2 From t=0 to t=1. From t=1 to t=2, the straight line continues to t=2. From t=2 to t=3, the straight line is at -(9/2). From 3 to 4 the graph is at -(9/2). From 4 to 5, the graph is at 2.
Using the fundamental theorem and the fact that P(t) when 2<=t<=4 is (1/2)t^2-3t+2 because c=2 when P(0)=2, we see that P(1)=0, P(2)=-2, P(3)=-(13/2), P(4)=-11, P(5)=-9
@& I believe this problem is covered in the Query, which you should submit.*@
5) f(x) is a function defined on the interval 0 <= x <= 7 with the following characteristics:
f '(x) = 1 for all x on the interval (0,2),
f '(x) = -1 for all x on the interval (2,3),
f '(x) = 2 for all x on the interval (3,4),
f '(x) = -2 for all x on the interval (4,6),
f '(x) = 1 for all x on the interval (6,7)
f(3) = 0
What is the integral of f ' (x) over the interval 0 <= x <= 7?
Describe your graph of f(x), indicating where it is increasing and decreasing and where it is concave up, where it is straight and where it is concave down.
What are the following values:
f(2)
f(3)
f(4)
f(6)
f(7)
Is the graph of f(x) continuous?
A: a) The integral of f’(x) between 0 and 7 is f’(7)-f’(0)=1-1=0
@& You are applying the Fundamental Theorem to a function which is not continuous on the region of integration. The fundamental theorem does not apply in this case.
If it did your conclusions would be valid. However you have to integrate the function separately on the intervals of continuity.*@
@& Think in terms of the area beneath the curve, but be careful about whether that area is above or below the axis.*@
b)f(x) is increasing 0<=x<=2, 3<=x<=4, 6<=x<=7
decreasing 2<=x<=3, 4<=x<=6
concave up 2<=x<=4, 4<=x<=7
concave down 0<=x<=3, 3<=x<=6
It’s never straight because the slope is never 0.
c) Using the fundamental therem, we find that
f(2)=1, f(3)=0, f(4)=2, f(6)=1, f(7)=2
d)yes
6) *6.1.10 Let f (x)= x^3 - 3x^2 and F'(x) = f(x). Sketch or graph f(x) (with a graphing calculator if necessary, though you should be able to construct the graph easily) and referring only to the graph of f(x), give the critical points, explain which are local maxima/minima, and which are neither. Also sketch a possible graph of F(x).
A: Critical points at x=0 and 3. L min x=2, L max x=0. Inflection Point x=1.
F(x) is a concave up parabola passing through x=0.
7) *6.1.12 Sketch and describe two functions F with F'(x) = f(x) (Sketch or graph f(x) with a graphing calculator) by only looking at the graph of f(x) = x^3 - 4x^2 + x + 6. In one let F(0) = 0 and the other F(0) = 1. Also identify local minima, maxima, and points of inflection.
A: Local min x=3
Local max x=2
Inflection point x=2.5
@& f ' would be 3 x^2 - 6 x; critical points would be at x = 0 and x = 2.
f '' would be 9 x - 6. Inflection point would be at x = 2/3, where the sign of the second derivative changes from positive to negative.
*@
8) *6.1.18 Let g' ={-x when 0 <= x <= 10, 4x when 10 <= x <= 15, x when 15 <= x <= 20, and -(1/2)x when x >= 20}. Sketch g' and given g(0) = 50 sketch the graph of g(x) and specify all critical and inflection points of g(x).
A: Inflection point x=10 and x=20
@& These are point of discontinuity, not inflection points.*@
Local min x=13
@& At x = 13 the function is increasing with slope 4. This isn't a locat min.*@
9) 6.1.22 Use a graph of 4cos(x^2) to determine where an antiderivative, F, of this function reaches its maximum on 0<= x<= 3. If F(1) =4, find this maximum obtained.
A: I don’t know how to find the anti derivative.
@& You need to do this by graphical reasoning.
The function reaches a relative max when its derivative changes from positive to negative.*@
10) Given the graph below, which depicts outflow vs. clock time and inflow vs. clock time for a reservoir:
When was the quantity of water greatest and when least? Describe in terms of the behavior of the two curves.
When was the quantity of water increasing fastest, and when most slowly? Describe in terms of the behavior of the two curves.
A: a) July greatest and Jan. 08 least.
between about march and july, the inflow is greater than the outflow showing that the greatest amount of water is accumulated in july and then decreases from there because the outflow is greater than the inflow and continues to be so up to June 08.
b) Little after april (may) fastest, Oct slowest.
the only time the inflow is greater than the out is between march and july and the highest point of the inflow is in may.
Oct is the slowest because the outflow is at its highest point above the inflow."
@& You are welcome to submit problems in this format, but first you should submit the Query for the assignment. With the feedback you get from the Query you can usually revise incorrect solutions before submitting them.
You appear to have a pretty good knowledge of derivatives and the Fundamental Theorem, but you don't always apply it correctly. So I think you're in good shape to start this course.
Check my notes, submit the Query, revise your solutions as necessary and I'll be glad to take another look.
Revisions should follow the guidelines given below.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#