rate of change / continuous piecewise fcn / etc

[sandman]

1) Find the instantaneous rate of change of g at x = 1, where g(x) = 4 cbrt[2x - 1].

2) Use implicit differentiation to find the derivative of -4y^3 + 3xy = -2x - 14

3) Given the following piecewise function:

. . . . . . . . ./ 3ax - 2bx^2 + 1, for x > 2
. .. .f(x) =< 2ax - 3, for x = 2
. . . . . . . . .\ 4ax^3 - 5b, for x < 2

Find values for "a" and "b" which make the function f(x) continuous everywhere.

4) A particle moves on the x-axis, with its position at any time t, t > 0, given by x(t) = 4t^3 - 18t^2 + 15t - 1. Find:

a) when the particle is at rest.
b) when the particle experiences its maximum velocity.
c) when the particle is moving to the left.

5) Find a cubic function of the form g(x) = ax^3 + bx^2 + cx + d that has a relative maximum value of 4 at -3 and relative minimum value of 2 at 0.

I'm sorry for the hodge-podge of questions. Any help on how to solve these would be super! I am not necessarily looking for answers; just the correct way to solve them.

Thank you!

 

[tkhunny]

Re: Assorted calculus troubles

How exactly woudl i solve for the instaneous rate of chane of g at x=1.
where g(x)=(4)3√(2x-1) (the 3 is 3rd root sorry for size)

Unfortunately, if you REALLY have no clue on any of these (as you seem to be indicating by your utter lack of demonstrated effort), you simply haven't been paying attention all semester and we will not be able to make your life any better.

You find the derivative of g(x) and evaluate it at x = 1. If you had read the definition of a derivative, even once, you should have known that. You can also apply the definition of the derivative, finding the limit of the defining expression as x approaches 1.

Personally, I'm not even a little encouraged. Start showing your work and I will change my mind rather rapidly.

 

[sandman]

Unfortunately, if you REALLY have no clue on any of these...we will not be able to make your life any better....

okay maybe i should have been more clear on this, i understand how to solve for the derivative and the instantaneous spead, but the follow up to that question askes for the g'(1) which i was under the impression was the exact same thing. I'm confused as to why they'd as the same question twice and ask to compare them.

 

[stapel]

Since you did not post whatever other parts there are to this exercise, nor did you post specifications regarding where you were confused, naturally the tutors could not respond to that. And since you did not show any work, naturally the tutors reasonably assumed that you couldn't even get started.

The tutors aren't trying to "sound smart"; they're merely responding to what you have posted. I apologize for any confusion.

Please reply with the full texts of the exercises under consideration, showing all of your work so far, and clearly stating where you are having difficulty.

Thank you.

Eliz.

 

[tkhunny]

I already gave you both methods to pursue. If you are asked to compare two results, quite obviously you must be required to find the result two different ways.

Finding g'(x) and evaluating at x = 1 is one way.

What is the Limit Definition of the Derivative? That is another way.

You may wish to devise your own methodology. The folks who invented derivatives did just that.

 

[galactus]

Re: Assorted calculus troubles

Also:
i'm not quite sure what to do with this problem the wording of it is confusing me
Find a cubic function of the form g(x)=ax^3+bx^2+cx+d that has a relative maximum value of 4 at -3 and relative minimum value of 2 at 0.

You can set up a system of equations and solve for a,b,c,d.

\(\displaystyle \L\\a(-3)^{3}+b(-3)^{2}+c(-3)+d=4\)
\(\displaystyle \L\\\underbrace{3a(-3)^{2}+2b(-3)+c=0}_{\text{derivative}}\)
\(\displaystyle \L\\a(0)^{3}+b(0)^{2}+c(0)+d=2\)
\(\displaystyle \L\\\underbrace{3a(0)^{2}+2b(0)+c=0}_{\text{derivative}}\)

Solve the system. I graphed it and it worked. Easy to make a mistake. Be careful.

Here's the graph. I left out the equation until you give it a go.

 

[pka]

This is a usual \(\displaystyle f(2 + ) = \lim _{x \to 2^ + } f(x)\) notation for the limit from the right at 2. Similarly, f(2-) is the limit from the left.

For #3, you will need f(2)=f(2+)=f(2-).

 

[sandman]

how is this for the cubic problem?

2/9x^3+x^2+2

 

[skeeter]

how is this for the cubic problem?

2/9x^3+x^2+2

no good ... f(-3) = 5, not 4.

 

[sandman]

2/9x^3+8/9x^2+2

 

[skeeter]

so ... what is this, guess and check?

2/9x^3+8/9x^2+2

again, no good ... f(-3) = 4 is not a relative maximum.

here you go ...

f(x) = ax³ + bx² + cx + d
f'(x) = 3ax² + 2bx + c

you were told ... relative max at (-3,4) and relative min at (0,2)

since (0,2) is on the curve, d = 2
now you have ...
f(x) = ax³ + bx² + cx + 2
(-3,4) is also on the curve ...
4 = a(-3)x³ + b(-3)² + c(-3) + 2
4 = -27a + 9b - 3c + 2
2 = -27a + 9b - 3c

max's and min's occur where f'(x) = 0 ...

0 = 3ax² + 2bx + c

0 = 3a(-3)² + 2b(-3) + c
0 = 27a - 6b + c

0 = 3a(0)² + 2b(0) + c ... c = 0

now you have two unknowns ...

2 = -27a + 9b
0 = 27a - 6b
-----------------
2 = 3b

b = 2/3

0 = 27a - 6(2/3)
a = 4/27

f(x) = (4/27)x³ + (2/3)x² + 2