# Thread: Derivatives, Abs. Extrema, Implicit Diff.

1. ## Derivatives, Abs. Extrema, Implicit Diff.

Hello guys, I have multiple questions today regarding several problems I have had problems working out. I will post the problem following how I have attempted to work it out. I have an exam on the topics of derivatives, absolute/relative extremas, implicit diff, relative rates later this week, so if you have any tips regarding these topics please share them.

1.) Find the derivative.
y = 23 ^ -x, I thought the answer to this would be ln(23)(23^-x), however in the answer key it states it as -23^-x, how is this so?

2.) 2nd Derivative of function f(x) = x/x+1, is the answer the same as the function itself? F''(x) = x/x+1?

3.) Abs. Extrema
f(x) = 1/x+2 [-4,1]
f'(x) = -x-2/(x+2)^2 ? x=0 x=-2, (0 , 1/2) ??? Is this how this is suppose to be worked out?

4.) Abs Extrema
f(x) = x^(4/3) - x ^(2/3)
f'(x) = 4/3x^1/3 - 2/3x^1/3, where do I go from here to solve this? I am confused as to how I would make some of these equations =0, any tips would be nice.

5--The last problem I have is a word problem related to business. I am not sure what it is exactly asking for in the problem however it states that there is a off-load oil docking facility 5 miles off shore. the nearest refinery is 8 miles easy of the facility. a pipe would cost 200k on land and 300k in water. Locate point B to minimize the cost. ------- So basically, there is a triangle drawn to show where the facility is and where the refinery is(on land), point B is suppose to be on land halfway from point A and the refinery. point A is the point on land that is 5 MILES from the facility off shore. Here I will just type out the problem to make more sense, I need more help with the problems above, this is just an extra however it will not be heavily tested.

Supertankers off-load oil at a docking facility shore point 5 miles offshore. The nearest refinery is 8 miles east of the docking facility a pipeline must be constructed connecting the docking facility with the refinery. the pipeline costs 300,000 per mile if constructed underwater and 200000 per mile if over land. Locate point B to minimize the cost of construction.
A triangle is drawn to show the facility point A(5 mile from docking) point B(from docking to a horizontal distance towards the refinery however not quite there.)

Thanks for all the help, I hope this post was not quite long however I waited until I was done with all of the problems before pointing out the ones I needed help with. Once again, thank you.

2. Any help works, on certain problems, still having problems figuring out how they work.

3. Anyone?

4. Originally Posted by cokeclassic
Hello guys, I have multiple questions today regarding several problems I have had problems working out. I will post the problem following how I have attempted to work it out. I have an exam on the topics of derivatives, absolute/relative extremas, implicit diff, relative rates later this week, so if you have any tips regarding these topics please share them.

1.) Find the derivative.
y = 23 ^ -x, I thought the answer to this would be ln(23)(23^-x), however in the answer key it states it as -23^-x, how is this so?
I presume you mean the answer was given as $-ln(23) 23^{-x}$, since just " $-23^{-x}$" is not correct. The "-" comes from using the chain rule. The derivative of $23^x$ is $ln(23) 23^x$ but since the exponent is "-x" you have to multiply by the derivative of -x which is -1.

2.) 2nd Derivative of function f(x) = x/x+1, is the answer the same as the function itself? F''(x) = x/x+1?
No, why would you think so? By the quotient rule, the derivative of f is $f'(x)= \frac{1(x+1)- x(1)}{(x+1)^2}= \frac{1}{(x+1)^2}$ and then, using the quotient rule again, $f"(x)= \frac{0(x+1)- 1(1)}{(x+1)^4}= \frac{x}{(x+1)^4}$. Did you forget to square in the denominator?

3.) Abs. Extrema
f(x) = 1/x+2 [-4,1]
f'(x) = -x-2/(x+2)^2 ? x=0 x=-2, (0 , 1/2) ??? Is this how this is suppose to be worked out?
First, your derivative is wrong. (I am assuming that you mean f(x)= 1/(x+2), not (1/x)+ 2 which, strictly speaking is what you wrote.) $f'(x)= \frac{0(x+2)- 1(1)}{(x+2)^2}= \frac{-1}{x+2}$.
An extermum occurs at one of three kinds of places.
(1) Where the derivative does not exist, which does not happen here.
This derivative does not exist at x=-2 but neither does the function value so we can ignore that.

(2) Where the derivative is 0. Since a fraction is 0 only where the numerator is 0, and this numerator, -1, is never 0, that alos does not apply.

(3) At the endpoints. f(-4)= 1/(-4+2)= -1/2 and f(1)= 1/3. The larger of those is 1/3 so the absolute maximum is 1/3 which happens at x= 1.

4.) Abs Extrema
f(x) = x^(4/3) - x ^(2/3)
f'(x) = 4/3x^1/3 - 2/3x^1/3, where do I go from here to solve this? I am confused as to how I would make some of these equations =0, any tips would be nice.
Once again, you have the derivative wrong. The derivative of $x^n$ is $nx^{n-1}$ and 2/3- 1= -1/3, not 1/3.

$f'(x)= (4/3)x^{1/3}- (2/3)x^{-1/3}= 0$ is the same as $(4/3)x^{1/3}= (2/3)x^{-1/3}$. Now multiply both sides of the equation by $3x^{1/3}$ and divide by 2: $2x^{1/3+ 1/3}= 1$ so $x^{2/3}= 1/3$. You solve that by taking the 3/2 power of both sides or, same thing, by taking the square root and then cube of both sides: $x= \pm\sqrt{1/27}$.

5--The last problem I have is a word problem related to business. I am not sure what it is exactly asking for in the problem however it states that there is a off-load oil docking facility 5 miles off shore. the nearest refinery is 8 miles easy of the facility. a pipe would cost 200k on land and 300k in water. Locate point B to minimize the cost. ------- So basically, there is a triangle drawn to show where the facility is and where the refinery is(on land), point B is suppose to be on land halfway from point A and the refinery. point A is the point on land that is 5 MILES from the facility off shore. Here I will just type out the problem to make more sense, I need more help with the problems above, this is just an extra however it will not be heavily tested.

A triangle is drawn to show the facility point A(5 mile from docking) point B(from docking to a horizontal distance towards the refinery however not quite there.)
First I am going to assume that you mean that the pipe would cost 200k per mile on land and 300k per mile off shore. Also B is NOT 'halfway' between A and the refinary- it is between them and the problem is to determin how far from A to B.

Let "x" be the distance from A to B- that will give you a right triangle with one leg "x" and the other 5. By the Pythagorean theorem, the straight line distance to the point where the pipe hits the coast is $\sqrt{x^2+ 25}$ miles. Since building the pipe off shore costs 300k per mile the cost of building that portion is $300000\sqrt{x^2+ 25}$. Since B is already distance x from A to the refinery, the distance remaining on land from B to the refinery is 8- x and the cost of building that will be 200000(8- x).

The total cost of the pipe line would be $C(x)= 300000\sqrt{x^2+ 25}+ 200000(8- x)$. That is what you want to minimize. You will want to differentiate it. You will probably find it simplest to write it as $C(x)= 300(x^2+ 25)^{1/2} + 200000(8- x)$.

Thanks for all the help, I hope this post was not quite long however I waited until I was done with all of the problems before pointing out the ones I needed help with. Once again, thank you.

5. $

f"(x)= \frac{0(x+1)- 1(1)}{(x+1)^4}= \frac{x}{(x+1)^4}
$
for this part, you took the derivative of 1/(x+1)^2, however I dont understand, what exactly do you do with the (x+1)^2? would it not be 2(x+1)(1)?