derivatives on graphs/Optimization problems

[JoeJ]

I have a couple questions pertaining to derivatives on graphs, and one on Optimization...

1. Let f'(x) = (x-3)(x-1) ^ 2. Answer the following questions about f(x)...

(a) On what intervals is f(x) increasing and decreasing? I'm thinking where f' is positive, at (3,infinity)

(b) On what intervals is f(x) concave up and concave down? Here, I'm concave up at thinking (-infinity, 1), and (3, +infinity), and concave down at 1,3).

(c) Which points on the graph of f are inflection points? Here, I'm thinking at (1,0), and (2,-1).

2. Find a positive number such that the sum of four times the square of the number and the number's reciprocal is as small as possible. (That is, 4x^2 + (1/x) is as small as possible).

 

[Gene]

You don't say how you got your answers but...
a) sounds good but you should say increasing and add decreasing elsewhere.

b) Right on the 1, close on the 3. Solve for f" unless you are limited to graphs.

c) If the constant C is zero, neither (1,0) nor (2,-1) is a point on f. At least one of them is wrong.
x = 2 isn't a solution for f" = 0

2. y = 4x^2 + (1/x)
y' = 8x-1/x^2 = 0
8x^3=1
x=.5

 

[JoeJ]

I got the answers by graphing, and looking at the graphs.

Ok, then for b) f" = (x -3)(x - 1)^2 = (x ^ 2 - 4 x + 3) ^ 2 = 2x^2 - 8x + 6. Not sure if this is the correct way or not.

and for c) you would then set f" = 2x^2 - 8x + 6 = 0? Wouldn't that be 2(x^2 - 4x + 3) = 0...but, that's the same as above.

 

[Gene]

Well, I get
f = (1/4)x^4 - (5/3)x^3 + (7/2)x^2 - 3x + C by integrating f'.
f' = x^3-5x^2+7x-3 by expanding (x-3)(x-1)^2
f" = 3x^2-10x+7 = (x-1)(3x-7) by differentiating f'
There are no initial conditions given so it is impossible to solve for C so I don't see how you can graph f unless you pick a C that you like and move the graph up or down to where ever you want it. Of course I'm cheating by doing it this way. I don't know how accurate you are supposed to be reading a graph so your answer of 3 may be good enough for 7/3.

Ok, then for b) f" = (x -3)(x - 1)^2 = (x ^ 2 - 4 x + 3) ^ 2 = 2x^2 - 8x + 6

I'm not sure what you did here but the highest exponent in f' is 3 so differentiating it can't have a higher exponent than 2 and you have an exponent of 4 in (x ^ 2 - 4 x + 3) ^ 2. I suspect a bad expansion of (x -3)(x - 1)^2.