You need to remember the pattern for calculating derivatives: d/dx xn = nxn-1. In this case we get df/dx = 4*6x5-3*5x4+2*1x0 = 24x5-15x4+2 (since anything to the zero power equals one and 2*1*1 = 2).
The product rule says that
Each of the factors, h and then g, gets a "turn" with the derivative while the other is left alone. In this problem, think of the first factor as (10x2-3x+2) and the second as (4x-2). Now we have
The quotient rule say that
| d/dx (h/g) = |
g h' - h g' g2 |
Which we've learned to remember with the "sing-song" phrase
lo d hi - hi d lo over (lo)2
Now we can calculate
| df/dx = |
(9x-2)(6x+2) - (3x2 + 2x +1)(9) (9x-2)2 |
The graph of the function f is
| y | = | f(x) |
| height | = | function value |
Because (0,0) is a point on the graph, I know that 0=f(0) (height = function value). But if we calculate, we find that f(0)=c, so c=0. We also know that 0=f(4), for the same reasons, and f(4)=16a+4b. That is...
We're on our way to finding the numbers a and b, but this is only a single equation and we have two variables. We need more information to pinpoint the exact numbers we want. The information we've not used so far is about the tangent line.
We know that the slope of the tangent line is told to us by the derivative, and that the slope will be 3 when x=4. That is, f'(3)=4. We need to find a formula for f'(x) if this information is to be at all helpful.
so we write f'(3)=4 as
6a + b = 4
Now we have two equations and two unknowns (a and b), so we can solve to get a=2 and b=-8.
To formulate the equation of a line, we need a point and a slope. The derivative tells us the slope of the tangent line, so we calculate
Substituting x=1, we see the slope of our tangent line is 11.
To get the reference point for our line, we use the equation given to us:
so when x=1, y=7. With a slope and point in hand, we use the point-slope formulation of a line and write
- The function s tells us the position of this object. To find the
position when time is t=0, find s(0)= -5 (i.e. 5 feet behind us).
- The object is stationary when its velocity is zero. Velocity
is the (instantaneous) change in position; the derivative of
position. So we calculate
v(t)= ds/dt = -6t2 + 42t - 60 I want to know when the velocity is zero, so I solve the equation
v(t)=0 -6t2 + 42t - 60 = 0
Notice that you can divide out a (-6) and, after that, the left-hand side of the equation factors into (t-5)(t-2). The equation is solved for t=2 and t=5, which are the times when the object is stationary.
- The object is moving forward when its velocity is postive. We
know the only times when v(t)=0 and that it's a continuous function,
so we need only check at various times t to see which intervals of
time have positive velocity.
- The object is moving backward when its velocity is negative. The
same technique mentioned above will work to find these intervals of
time. I've put a table below....
When t=... v(t) is so the object's moving... 0 negative backward during (- 
,2)3 positive forward during (2,5) 6 negative backward during (5, ) - The object is speeding up when its acceleration and velocity are
in the same direction and is slowing down when they are in opposite
directions. We can identify these times by comparing the parity of
velocity to that of acceleration. To work with acceleration, however,
we must first have a formulation it in hand. We've said in class that
acceleration is the change in velocity. That is, the derivative of
velocity, so we calculate
a(t) = dv/dt = -12t+42 It's not hard to see that a(t)=0 when t=21/6 , which is a little less than 3. This is the only time when a(t)=0, so to check whether it's positive or negative we need only check a pair of points; one greater than 3 and one less than 3. We find that a(t) is positive if t<21/6 and negative if t>21/6 . Now, to finish the question, it's probably best to make a chart...
Per the above comments, we can now see that...
- the object is speeding up during
(2,21/6)
(5, )
- the object is slowing down during
(- ,2)
(21/6,5)
- the object is speeding up during
(2,21/6)
To answer any of these questions, we need to have a formulation for the height of the ball as a function of time. We know the height function looks like
where v0 is the initial velocity and h0 is the intial height. For this problem, the height function is then
- The ball hits the ground when the height is zero. That is, when
h(t)=0. Using the quadratic formula we find that this happens when
t=4.6538 or t=-0.0288. Clearly, the ball will hit the ground
after we throw it, so we're interested in the positive answer.
- The impact velocity is the "downward" velocity - how fast the
height is changing at the instant the ball hits the ground. To find
the velocity, we take the derivative of the height function.
v(t) = -32t + 10 Now we have something that tells us about velocity, and we know that we're interested in the velocity at time t=4.6538 (that's when the ball hits the ground), so we calculate v(4.6538) = -138.92. That is, 138.92 feet per second, downward (remember, the parity of velocity tells us direction).
- We know that, when the ball reaches its highest point, the
velocity will be zero. We've got a function that tells us the
velocity, so let's use it to find out when the velocity is zero. That
is, solve v(t)=0. This happens when t=0.3125 seconds.
- The function h tells us about the height of the ball, and we know the time at which it reaches the greatest height, so we simply calculate h(0.3125)=301.5625 feet (we didn't get much oomph on that toss).
- The rate of change is told to us by the derivative, so we
calculate
dP/dt = 4 + 5/2 t3/2 - 1/2 t-1/2 To answer the question, we take t=9 and find that P'(9)= 71.3333
- The percentage rate of change is calculated by finding the quotient P'(9)/P(9) and then multiplying by 100. We already have the number P'(9), so we need only calculate P(9)=5276. The percentage rate of change is 71.3333/5276*100 = 1.35.
I begin with a note on exponents: Recall that x5/2 the same as
(
)5.
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- "Rate of change" is a key phrase that should set off alarm kaxons
in your head. The derivative tells you about the rate of
change, so the first part of this exercise will be answered by using the
derivative.
I'll write D(p) for the demand, as a function of price. Note also that price is a function of time, t. So we're looking at D(p(t)) - a composite function! To find the derivative of D(p(t)) with respect to time, the chain rule tells us that
d/dt D(p(t)) = D'(p(t)) p'(t) In particular,
d/dt D(p(10)) = D'(p(10)) p'(10)
That was the hard part. Now we just go and get these numbers and put them together according to the formula.
- p(10) = 0.83
- D'(p) = -4260p/(1+p2)2 so D'(0.83)=-1222.13
- p'(t) = 0.002t - 0.01 so p'(10)=0.01
Therefore, d/dt D(p(10)) = D'(p(10)) p'(10) = (-1222.13)(0.01) = -12.2213.
- p(10) = 0.83
- To find the percent rate of change, we need only divide our
answer from part (a) by D(p(10)) and then multiply by 100 (according
to the formula for percent rate of change).
D(p(10)) = D(0.83) = 1261.1759 so the percent rate of change is 100*-12.2213./1261.1759 = -0.969
The answer is zero.
To find the second derivative of h, take the derivative of h'(x). The answer is
The graph will be increasing exactly when the slope of the tangent line is postive and decreasing exactly when the slope of the tangent line is negative.
The slope of the tangent line is told to us by the derivative, so we need to find h'(x) and will use the Chain Rule in doing so.
- It's easy to check that h'(1)>0, so the graph is increasing at
x=1.
- It's easy to check that h'(-1)<0, so the graph is decreasing at x=-1.
You should be able to find the following important characteristics of the graph without much difficulty:
- y-intercept: (0,1)
- h'(x) = x2 + 2x - 3
- h'(x)=0 when x= 1 or x=-3
... so the graph may have a maximum or minimum at either of these points.
- h'(x)<0 in the interval (-3,1)
... so the graph is decreasing on that interval.
- h'(x)>0 everywhere else
... so the graph is increasing on that interval.
Your graph should look something like this...
You should be able to find the following important characteristics of the graph without much difficulty:
- y-intercept: (0,0)
- The derivative is
h'(x) = -3x2+15
(x2+5)2
Finding when h'(x) is zero is slightly more tricky this time, but still within our grasp. Note that h'(x) is a fraction. The only possible way a fraction could be zero is if the numerator is zero. So to find out when h'(x) is zero, we need only find out when -3x2+15=0, which happens when x=2.2361 and when x=-2.2361.
As above, the values of x where h'(x)=0 may be local maxima or minima.
To check for increase and decrease of the function, we want to know the parity of h'(x). This looks tricky, but it's not; for two reasons. First of all, we know when h'(x)=0 and, since it's a continuous function, those are the only opportunities h' has to change from positive to negative. That means we need only check a few different values of x.
On the other hand, you might notice that the denominator of h'(x) is always positive, so it will never affect the parity. We can make our decision based solely upon the parity of the numerator.
What you'll find is the following:
- h'(x)>0 on the interval (-2.2361,2.2361)
... so the graph should be increasing on that interval
- h'(x)<0 everwhere else
... so the graph of h should be decreasing everywhere else.
Your graph should look something like this...
- To determine whether a function is increasing or decreasing you must
know about its derivative so the first thing I do is take the
derivative of h.
h'(x) = 6x2 + 10x -8 I care about whether the derivative is positive or negative. To find out, I'm going to look for the times when h'(x)=0, and check between them. That is, solve the equation
6x2 + 10x -8 = 0 I used the quadratic formula to find that x=0.5907 or x=-2.2573. Now I check h'(x) at some arbitrary, but well chosen values of x.
When x=... ...h'(x) is... ...so the graph of h is... -3 positive increasing on (- ,-2.2573)0 negative decreasing on (-2.2573,0.5907) 1 positive increasing on (0.5907, )You should understand the logic here: I know that h'(x) is a continuous function (it's a quadratic polynomial), so the only time it could possibly switch from positive to negative, or vise versa, is when it's zero.
For example, could h'(10) be negative? No. I know that h'(1) is positive. If h'(10) were negative, I'd have a positive value at x=1 and a negative value at x=10, so this continuous function would absolutely have to have a root somewhere in the interval (1,10). But I know where all the roots of the function are and there aren't any in (1,10), so h'(10) is certainly positive.
- To determine concavity, you have to look to the second
derivative: h''(x). We calculate
h''(x)=12x+10. Again, we find where this function is zero. It's easy to see that h''(x)=0 when x = -0.8333. Now, just like in part (a), we check the parity of h'' at selected values of x.
When x=... ...h''(x) is... ...so the graph of h is... -1 negative concave down on (- ,-0.8333)0 positive concave up on (-0.8333, )
The chain rule tells us that to find the derivative of f(g(x)) we do the following:
In particular...
d/dx f(g(-1)) = f'(g(-1)) g'(-1)
Clearly, we need to know g'(-1). We also need to know f'(... but where? We evaluate at g(-1). It appears, from the graph of the function, that g(-1)=-2 so
To estimate this product, I estimate each piece.
(Remember, this is an estimate only, so there's no need to get really picky with the numbers. A reasonable, educated guess will do.)
To get g'(-1), draw the tangent line to the graph of g at x=-1.
The tangent line appears to pass through the points (-1,-2) and (-2,-4) - more or less - which means it has a slope of about 2. That is, g'(-1) = 2 - roughly speaking.
To get f'(-2), draw the tangent line to the graph of f at x=-2.
The tangent line appears to pass through the points (-3.8,0) and (-2,-1), which means it has a slope of about -0.55. That is, g'(-2) = -0.55 - roughly speaking.
Put these pieces of data together now, and you have the answer. We estimate
...
- In this case we see a 0/0 as the limit
value, which is one of the indeterminate forms to which we can apply
L'Hopital's Rule.
The derivative of the numerator is 3x2+6x-1.
The derivative of the denominator is 3x2+18x+26.We take the limit as x tends toward -3 of the quotient (3x2+6x-1) / (3x2+18x+26) and get the number -8, which is the answer.
- In this case we do not need to - and CANNOT - use L'Hopital's
rule, since the limit value is NOT one of the indeterminate forms.
The limit value is calculated directly as 1/4.
- When we calculate the limit value we see
0/0, which is one of the indeterminate forms, so
we apply L'Hopital's rule.
The derivative of the numerator is 6x2-10x+4.
The derivative of the denominator is 3x2-2x-1.We take the limit as x tends to 1 of the quotient (6x2-10x+4) / (3x2-2x-1) and, again, see the indeterminate form 0/0. So again we apply L'Hopital's Rule...
The derivative of the numerator is 12x-10.
The derivative of the denominator is 6x-2.We take the limit as x tends to 1 of the quotient (12x-10) / (6x-2) and get the number 1/2 , which is the answer.