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)
PLEASE HELPP THIS IS URGENTTT! thank u!! appreciated!
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
For simplicity, set . Then
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
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 ........
Ha ha, Well, I guess the typo:
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.Originally Posted by lemontea
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.