Recent content by k3k3

Homework Statement Generalize: For arbitrary 0 < p < 1, show that the method giving a and b produces the minimum length interval. Hint: It might be helpful to use local extrema for the inverse function of the distribution function. Homework Equations The method is is talking about is...

Homework Statement Assume the function f defined by f(x)=5x+sin(πx) is strictly increasing on ℝ. Find (f^{-1})'(10) Homework Equations Let I and J be be intervals and let f:I->J be a continuous, strictly monotone function. If f is differentiable at c and if f'(c)≠0, then (f^{-1}) is...

h'(c) must be 0 too since c is in (a,b)

h'(c) = [h(b)-h(a)]/(b-a) ≠ 0

That the function has different values at the end points of the interval.

Let me try this again, then... I want to show that if f'=g' on (a,b), then there exists a constant k such that f(x)=g(x)+k for all x in [a,b] Let f and g be two continuous functions that are defined on [a,b] and differentiable on (a,b). Assume f'(x)=g'(x) for all x in [a,b] Let...

What would bother me that is, when f'(x)=g'(x), is that h(a) is h(b). And when h'(x) = 0 for all x in (a,b), then h(x) is constant.

MVT tells me that there is a point in the interval (a,b) such that h'(c) = [h(b)-h(a)]/(b-a) So f'(c)-g'(c) = [h(b)-h(a)]/(b-a) and f'(x) = g'(x) for all x in [a,b] So 0 = [h(b)-h(a)]/(b-a) => h(b)=h(a) making it a constant function.

So choose x,y in (a,b) arbitrary so that h(x)-h(y) = 0?

If h(x)=f(x)-g(x), then h'(x)=f'(x)-g'(x) and, since f'=g' by assumption, f'(x)-g'(x)=0 for all x in (a,b). By MVT, there is a c in (a,b) so that h'(c)=h(b)-h(a) = 0 or [f(b)-g(b)-f(a)+g(a)]/(b-a) = 0 => f(b)-f(a)=g(b)-g(a) ? If h'(x) = 0 for all x in (a,b), then h is constant. Is this...

Homework Statement Let f and g be two continuous functions that are defined on [a,b] and differentiable on (a,b). Claim: If f'=g' on (a,b), then there exists a constant k such that f(x)=g(x)+k for all x in [a,b] Homework Equations The Attempt at a Solution Let h:[a,b]->R be...

So I just need a function whose values are between -pi/2 to pi/2?

I've thought about that, but I have had no luck in doing so.

By bounded I mean it is bounded above and below, yeah. I can't think of a linear function that would scale with the arctan. How about 1/x when x isn't an integer?

How about arctan(x) if x is in (n, n+1) and arctan(x)+1/x if x is an integer?