# Math Help - Functions question!! URGENTT..!!

1. ## Functions question!! URGENTT..!!

Let f(x)=2x^4-x^3-2x^2+14 and g(x)= x^4-x^3+2x^2+10

1)Show, using the intermediate theorem that there is a point in the [0, 1] interval where f(x) and g(x) are equal

2) Find the value x E [0, 2] for which f(x)=g(x)

2. Originally Posted by lemontea
Let f(x)=2x^4-x^3-2x^2+14 and g(x)= x^4-x^3+2x^2+10

1)Show, using the intermediate theorem that there is a point in the [0, 1] interval where f(x) and g(x) are equal

2) Find the value x E [0, 2] for which f(x)=g(x)

1) Consider the function h(x) = f(x) - g(x). If you can show that h(a) < 0 and h(b) > 0 for some values a and b in the interval [0, 1], then you know that h(x) = 0 for some x in [0, 1].

2) What have you tried?

3. 1.) I am a little confused with what you stated in #1...><'''..

2.) No I haven't tried anything yet, because i dont know what the questioni s asking for.

4. Originally Posted by lemontea
1.) I am a little confused with what you stated in #1...><'''.. Mr F asks: Where, exactly, are you confused?

2.) No I haven't tried anything yet, because i dont know what the questioni s asking for. Mr F says: The question is crystal clear. It's asking for a value of x in the interval [0, 2] that makes f(x) equal to g(x).
..

5. 1.) Is what you stated what i should write in my answer...how do i show that h(a) < 0 and h(b) > 0??...

2.) To make f(x)=g(x), im thinking i have to make the two equations equal to eachother, so

2x^4-x^3-2x^2+14 = x^4-x^3+2x^2+10

and then i solve for x???...would this be correct?

6. Originally Posted by lemontea
1.) Is what you stated what i should write in my answer...how do i show that h(a) < 0 and h(b) > 0??...

2.) To make f(x)=g(x), im thinking i have to make the two equations equal to eachother, so

2x^4-x^3-2x^2+14 = x^4-x^3+2x^2+10

and then i solve for x???...would this be correct?
I note the following: There is no value c in [0, 1] such that f(c) = g(c). There is a typo in the problem. Perhaps the interval is supposed to be [0, 2] as is in the second problem?

Additionally, I didn't respond to this one because I can't find a way to apply the intermediate value theorem to it. mr fantastic's function h(x) is never negative, so you cannot find points a and b such that h(a) < 0 and h(b) > 0.

For 2), you have the right approach. The equation is a "biquadratic," an equation of the form
$ax^4 + bx^2 + c = 0$

For simplicity, set $y = x^2$. Then
$ay^2 + by + c = 0$

You can solve this for y in any way that you please. Then take the possible solutions for y and take the square root of them. That gives the solutions for x.

-Dan

7. topsquark, you are right, there is a typo in #1...

can anyone else help me with #1???

8. Originally Posted by lemontea
topsquark, you are right, there is a typo in #1...

can anyone else help me with #1???
I assume the typo was the interval and that it's meant to be [0, 2]?

I which case everything I said still holds! - just replace with the correct interval!

Let's see now ..... some value a in [0, 2] such that h(a) < 0 .... hmmmmmm. Let's try an endpoint, say 2:
h(2) = -12 < 0. Bingo!

Now for some value b in [0, 2] such that h(b) > 0 ....... I wonder if the other endpoint will work .....?

h(0) = 4 > 0. Aha!

Therefore, by the intermediate value theorem, there must exist an x in [0, 2], say x = c, such that h(c) = 0. Therefore ........

9. Originally Posted by topsquark
I note the following: There is no value c in [0, 1] such that f(c) = g(c). There is a typo in the problem. Perhaps the interval is supposed to be [0, 2] as is in the second problem?

Additionally, I didn't respond to this one because I can't find a way to apply the intermediate value theorem to it. mr fantastic's function h(x) is never negative, so you cannot find points a and b such that h(a) < 0 and h(b) > 0.
[snip]
Well, that's what I get for making a perfectly good suggestion (and it still is, by the way - just replace the interval) without bothering about the details.

10. Originally Posted by mr fantastic
Let's see now ..... some value a in [0, 2] such that h(a) < 0 .... hmmmmmm. Let's try an endpoint, say 2:
h(2) = -12 < 0. Bingo!
$h(x) = f(x) - g(x) = (2x^4 - x^3 - 2x^2 + 14) - (x^4 - x^3 + 2x^2 + 10) = x^4 - 4x^2 + 4$

How did you get $h(2) = -12$? I get $h(2) = 4$.

-Dan

11. Originally Posted by topsquark
$h(x) = f(x) - g(x) = (2x^4 - x^3 - 2x^2 + 14) - (x^4 - x^3 + 2x^2 + 10) = x^4 - 4x^2 + 4$

How did you get $h(2) = -12$? I get $h(2) = 4$.

-Dan
Ha ha, Well, I guess the typo:

Originally Posted by lemontea
topsquark, you are right, there is a typo in #1...

can anyone else help me with #1???
wasn't the interval. (And I guess I can't read my own writing - admittedly done on the back of a napkin). Pity we were told that there was a typo but not what the typo actually was ..... Pretty hard to offer useful advice when the question has mistakes in it.

Be that as it may, I'm done with this thread - looking like a chump twice (my monthly quota) in one day is where I draw the line in the sand.