Prove that there is some number x such that ?

[LilTaru]

Homework Statement

Prove that there is some number x eR such that:

a) x¹⁷⁹ + 163/(1 + x² + sin²x) = 119

b) sinx = x - 1

Homework Equations

The Attempt at a Solution

(a) just looks terrifying to me... especially the x¹⁷⁹! I don't know how they want me to prove it... Looks impossible to me! Any guidance would be appreciated!

 

[Char. Limit]

First, make sure that f(x) is continuous. Then, ensure that it has a point below 119 and a point above 119. Then, by some theorem or another, it must contain 119.

 

[LilTaru]

Would the same be for part (b) as well? And how do you prove a function is continuous?

 

[Char. Limit]

Would the same be for part (b) as well? And how do you prove a function is continuous?

Erm...

I can't quite remember how to prove if a function is continuous (I suppose if it's not, there would be points at which it didn't exist, i.e. division by zero or something), but the process for (b) will be the same.

 

[LilTaru]

Now would f(x) = x¹⁷⁹ + 163/(1 + x² + sin²x)
or f(x) = x¹⁷⁹ + 163/(1 + x² + sin²x) - 119? I was thinking the first and then find x so that f(x) = 119?

 

[jdc15]

By the intermediate value theorem, if we can find a value of x for which f(x)<119 and a value of x for which f(x)>119, and f(x) is continuous, then somewhere along that interval f(x)=119. That's how the theorem works.

To prove continuity, there are a bunch of rules. f(x)/g(x) for example is continuous where both f(x) and g(x) are continuous and g(x) is not zero.

 

[LilTaru]

Oh! Think I got it... to prove continuity can I not say F(x) = f(x) + g(x) where f(x) = x¹⁷⁹ and g(x) = 163/(1 + x² + sin²x)? Then prove that since each is continuous F(x) is continuous since it is the sum of two continuous functions?

 

[tjackson3]

The way I would approach this is the following:

Let f(x) = x^{179} + 163/(1+x^2+\sin^2(x)) - 119. In order to show what you wanted to show, it suffices to show that there exists some x_0 such that f(x_0) = 0. We can do that by showing that f is positive somewhere and negative somewhere else; then if f is continuous, the intermediate value theorem says that f has a zero somewhere in the middle. The thing that pops out to check about continuity in this is the denominator of that rational expression: 1+x^2+\sin^2(x). We need to make sure that it is never zero. I'll leave that part to you, but it's pretty simple when you consider what would need to happen in order for it to be zero.

Now we just need to find a positive and a negative number for f. If x << 0 (i.e. hugely negative), then x^{179} is hugely negative. The rational expression is positive and small, and the -119 would certainly nullify it. Therefore for very negative x, f is negative (in fact, x = -2 would work). The same logic applies for x >> 0 (or even x = 2). So we have a continuous function which is negative on one part of its domain and positive on the other; therefore, it must be zero in between.

 

[LilTaru]

It works for part (a), but I don't know how to formulate f(x) for part (b)...
Would it be f(x) = sinx - x + 1 and find when it equals 0?
Or f(x) = sinx + 1 and find when it equals x?
Or f(x) = sinx - x and find when it equals 1?

 

[Dick]

It works for part (a), but I don't know how to formulate f(x) for part (b)...
Would it be f(x) = sinx - x + 1 and find when it equals 0?
Or f(x) = sinx + 1 and find when it equals x?
Or f(x) = sinx - x and find when it equals 1?

Go with f(x)=sin(x)-x+1. What's f(0)? What's f(pi)?