calculus bc problem....?

Question

Let f(x) be a function that is twice defined and twice differentiable for all real numbers x and that has the following properties:
1. f(0) = 2
2. f ' (x) > 0
3. The graph of f(x) is concave up for all x > 0 and concave down for all x < 0

let g be the function defined by g(x)= f(x^2)

a. find g(0)
b. find the x-coordinates of all minimum points of g.
c. were is the graph of g concave up?

please help me. i have no clue.

Answer 1

lets take a look at f(x) = x^3, to adjust for the propert f(0)=2 we write f(x)=x^3+2 => f(0)=2

Verify f'(x)>0:
f'(x)=3x^2 >0 for all reals

Verify f(x) is concave up for all x > 0 and concave down for all x < 0
f''(x)=6x =>f''(x)>0 if x>0 and f''(x)<0 if x<0

Define g(x)=f(x^2)=x^6+2

a.) g(0)=x^6+2=0

b.) to find minimums we need to have g'(x)=0 and be concave up on either side of x. So we first derive g'(x)
g'(x)=6x^5
Well that makes it pretty easy because g'(x)=0 only at x=0
then verify that it is conave up on both sides:
g''(x)=30x^4
Even easier, g(x) is always concave up except at x=0 which is the single minimum of g

c.) as stated in our exploration for g''(x) is always positve except at zero, however it generally can still be stated g is concave up for all x

Answer 2

OK, as a disclaimer, I'm not sure on this. What is a function that is "twice defined", for instance? But let me offer some observations.

First off, looking at condition number 3, "concave up for all x>0 and concave down for all x<0", makes me think of x³.

But x³ would have f'(x) = 0 at 0, which would go against #2...you sure #2 isn't "f'(x) >= 0"?

In any case, if f(x) = x³ + 2 (to match #1), then

1) g(0) = (0²)³ + 2 = 2
2) g(x) = (x²)³ + 2 = x^6 + 2
g'(x) = 6x^5 = 0, so the minimum would be at x=0.
3) g''(x) = 30x^4, which is always greater than 0, so it's concave up at all points.

Again, I'm not sure, since my guess violates #2.

Answer 3

Consider f"(x) = 6ax, a>0

Then f"(x) > 0 for x > 0 and f"(x) < 0 for x < 0 ... so condition 3 is met

f'(x) = 3ax² + b >0 if b > 0 ... so condition 2 is met

f(x) = ax³ + bx + c. If c = 2 then f(0) = 2 ... so condition 1 is met

Thus f(x) = ax³ + bx + 2 is ONE such function

So g(x) = f(x²) = ax^6 + bx² + 2 (a, b > 0)

So g(0) = 2

g'(0) = 6ax^5 + 4bx
= 2x(3ax^4 + 2b)
= 0 for stationary points
So x = 0 or 3ax^4 + 2b = 0
So x = 0 and there are no real solutions to 3ax^4 + 2b = 0 as a, b > 0

So minimum at (0, 2)

g"(x) = 30ax^4 > 0 for all x except x = 0

Therefore concave up for all x except x = 0

In general f(x) = A_0*x^(2n - 1) + A_1*x^(2n - 3) + A_2*x^(2n - 5) + ... + A_(n - 2)*x³ + 2 will meet these conditions provided all coefficients are positive

f(0) = 2
f'(x) = (2n - 1)A_0*x^(2n - 2) + (2n - 3)A_1*x^(2n - 4) + (2n - 5)A_2*x^(2n - 6) + ... + 3A_(n - 2)*x²
> 0 for all x
f"(x) = (2n - 1)(2n - 2)A_0*x^(2n - 3) + (2n - 3)(2n - 4)A_1*x^(2n - 5) + (2n - 5)(2n - 6)A_2*x^(2n - 7) + ... + 6A_(n - 2)*x
= x [(2n - 1)(2n - 2)A_0*x^(2n - 4) + (2n - 3)(2n - 4)A_1*x^(2n - 6) + (2n - 5)(2n - 6)A_2*x^(2n - 8) + ... + 6A_(n - 2)]
> 0 for x > 0
and < 0 for x < 0

Whence
g(x) = f(x²) = A_0*x^2(2n - 1) + A_1*x^2(2n - 3) + A_2*x^2(2n - 5) + ... + A_(n - 2)*x^6 + 2, all A_k > 0

g'(x) = 2(2n - 1)A_0*x^(4n - 3) + 2(2n - 3)A_1*x^(4n - 7) +2(2n - 5) A_2*x^(4n - 11) + ... + 6A_(n - 2)*x^5

= 2x^5 [(2n - 1)A_0*x^(4(n - 2)) + (2n - 3)A_1*x^(4(n - 3)) +(2n - 5) A_2*x^(4(n - 4)) + ... + 3A_(n - 2)]

> 0 for x > 0
< 0 for x < 0
= 0 for x^5 = 0 or (2n - 1)A_0*x^(4(n - 2)) + (2n - 3)A_1*x^(4(n - 3)) +(2n - 5) A_2*x^(4(n - 4)) + ... + 3A_(n - 2) = 0

ie x = 0, 0, 0, 0, 0 and the other equation is positive definite ie no real roots since each term > 0 for all x (its an even function with a minimum at 3A_(n - 2))

So there is a minimum at (0, 2) only

And g"(x) = 2(2n - 1)(4n - 3)A_0*x^(4(n - 1)) + 2(2n - 3)(4n - 7)A_1*x^(4(n - 2)) +2(2n - 5)(4n - 11) A_2*x^(4(n - 3)) + ... + 30A_(n - 2)*x^4]
> 0 for all x > 0

Now this is only a polynomial solution The function f(x) = -arctan(x) + 2 also obeys your conditions so the whole solution is very open -ended

In reality f(x) = any even function with all coefficients positive and f(0) = 2 will satisfy your conditions

Answer 4

#a. g(x) is even function and for x<0 mirrors f(x) for x>=0.
#b. Since f'(x) > 0 f(x) is monotonically increasing so is g(x) for x>0.
From #a. and #b. follows that g has minimum at x=0 (answer to b.) which is g(0) = f(0) = 2 (answer to a.). Since f(x) is concave up for x>0 g(x) is also concave up for x>0. From #a. follows that graph is concave up for x<0 as well.

Answer 5

g(0) = f(0) = 2 (x^2 = 0, f(x^2) = f(0).

F(x) is concave up for all x>0; since x^2 is always > 0, then f(x^2) is always concave up. The minumum must then be at x = 0.

Answer 6

Check the given information. E.g. I don't think you really mean that "the function is twice defined." If the definitions conflict, which one will you work with? Or do you simply mean, "It really, really has the following properties!"?

Answer 7

best of luck to you, i'm two weeks from halfway through Calc B/C, and it is a toughh class. Right now we're learning about the Disc and Washer Methods.