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Jabatheblob1

Member

Joined: 2015-03-01

Posts: 856

MATH HELPS

The following two equations intersect ONCE between the listed domain.
f(x) = 3(tan(x))^4 + 2k
g(x) = -(tan(x))^4 + 8k(tan(x))^2 + k

0 <= x <= 1
0 < k

What is K? And how did you get it.

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Pingohits

Banned

From: aids lizard

Joined: 2015-02-15

Posts: 7,591

Re: MATH HELPS

i did it the nub way and graphed it with my ti-84, then used guess and check

k=0.25

this obviously doesnt help you besides the answer, since i didnt use a preferable method
wait a few and maybe someone else will give a better explanation

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Stagecrew

Member

Joined: 2015-02-15

Posts: 289

Re: MATH HELPS

If you set k high enough (k>2 for example) then they will intersect once in the interval, but let's ignore this case for now. Take u=(tanx)^2, turning f into 3u^2+2k and g into -u^2+8ku+k. We want these transformed functions to be tangent in the interval 0<=u<=(tan1)^2 - set the discriminant of f-g to zero and make sure the root of f-g is in that interval. The discriminant of f-g is 64k^2-16k, having roots at 0 and 0.25. Discard 0, it's not in the range. Checking at k=0.25, the root of f-g is in the right interval.

Now for the higher k, we want the transformed version of f-g to have a root at (tan1)^2 for the lower bound. Solving for this gives k=1.279, so if k>=1.279 it also works.

Jabatheblob1

Member

Joined: 2015-03-01

Posts: 856

Re: MATH HELPS

If you set k high enough (k>2 for example) then they will intersect once in the interval, but let's ignore this case for now. Take u=(tanx)^2, turning f into 3u^2+2k and g into -u^2+8ku+k. We want these transformed functions to be tangent in the interval 0<=u<=(tan1)^2 - set the discriminant of f-g to zero and make sure the root of f-g is in that interval. The discriminant of f-g is 64k^2-16k, having roots at 0 and 0.25. Discard 0, it's not in the range. Checking at k=0.25, the root of f-g is in the right interval.

Now for the higher k, we want the transformed version of f-g to have a root at (tan1)^2 for the lower bound. Solving for this gives k=1.279, so if k>=1.279 it also works.

I wasn't able to follow you very well but this seems about right. This was a no calculator part of a calculus test though. So the first method you seemed to used seems to use no calculator which would work.

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Abelysk

Guest

Re: MATH HELPS

What math level does this come out of? Pre-cal?

Jabatheblob1

Member

Joined: 2015-03-01

Posts: 856

Re: MATH HELPS

What math level does this come out of? Pre-cal?

Lower level college calc.

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Stagecrew

Member

Joined: 2015-02-15

Posts: 289

Re: MATH HELPS

If you set k high enough (k>2 for example) then they will intersect once in the interval, but let's ignore this case for now. Take u=(tanx)^2, turning f into 3u^2+2k and g into -u^2+8ku+k. We want these transformed functions to be tangent in the interval 0<=u<=(tan1)^2 - set the discriminant of f-g to zero and make sure the root of f-g is in that interval. The discriminant of f-g is 64k^2-16k, having roots at 0 and 0.25. Discard 0, it's not in the range. Checking at k=0.25, the root of f-g is in the right interval.

Now for the higher k, we want the transformed version of f-g to have a root at (tan1)^2 for the lower bound. Solving for this gives k=1.279, so if k>=1.279 it also works.

I wasn't able to follow you very well but this seems about right. This was a no calculator part of a calculus test though. So the first method you seemed to used seems to use no calculator which would work.

If you can't use a calculator, here's a slightly different way to approach the second part: we can express it in exact form in terms of (tan1)^2. We want f-g=4u^2-8ku+k to be 0 at u=(tan1)^2 so 4tan^4(1)=(8tan^2(1)-1)k meaning that k>=4tan^4(1)/(8tan^2(1)-1)=1.279 works. You can think of the change tan^2x->u as just plugging in values for tan^2x instead of x for your function - it's still the same thing, you just plug in different stuff. It's nice to make the substitution though because it reduces to a quadratic which is easier to deal with.