Is f(x) =sin x bijective?
No, certainly not. Sine function is not bijective function.A function is bijective if and only if it is onto and one-to-one. Let us first check, whether it is injective(one-to-one)According to horizontal line test, a curve is injective(one-to - one) only if a horizontal line cuts the curve only once.But as it is clear from above image, horizontal line cuts Sine wave in more than one points, so it is not inijective. And not bijective.But there is special type of Sine function, called, principal Sine function, which is a inijective function. When domain of Sine function is restricted to [math][-\frac{\pi} {2}, \frac{\pi} {2}][/math], it is calld principal Sine function(curve) . The Sine function is inijective therein, as a horizontal line cuts principal Sine function only once. Also it is onto function therein. So principal sine function is bijective. Inverse of principal Sine function can be found and is called inverse Sine function.So general Sine function is not bijective, but principal Sine function is bijective.
If fog= sinx^2 and gof = sin^2 x, what is f(x) and g(x)?
fog(x) = f(g(x)) = sin(x²)gof(x) = g(f(x)) = (sin(x))²Does this notation make it easier to see the result? Try your hand at it for a few more seconds again before scrolling further down.The working method you might want to use here would be to outline what things you're doing to x in each of the composite functions.fog(x) means first you do "g" to an x, then you do "f" to the result. In this case, we square a number then take the sine of the squaregof(x) means first you do "f" to an x, then you do "g" to the result. In this case, we take the sine of a number, then we square the sineCan you see how it makes sense for f(x) = sin(x) and g(x) = x²?
How we can show sin(x) / x is convergent?
Try using the integral test. Look at the integral of sin(x)dx/x for x = 1 to infinity. This integral is known to converge. (Using complex analysis the integral of sinx /x from 0 to infinity is known to converge to pi/2.)
How to show ∫(0 to 2π) f(x) cos(x) dx ≥ 0?
Scythian's function f(x) = 1 / (1 + (x-π)²) isn't convex on (0,2π). I think one fairly straightforward approach would be to take two cases, the first of which is easily disposed of: i) f monotone ii) f attains a minimum at (c,f(c)). (It might be easier to split case i into the separate cases increasing and decreasing, but since cosx is symmetric on the interval, it seems unnecessary.) If I have more time, I may come back and address this more explicitly; however, we just had a baby today, so I'm kinda doubting it! § ** CLARIFICATION: ** While Steiner's definition is correct, it's also a bit more abstract than necessary in this context. A convex C² function on the reals just means "concave-up" in the first-semester calculus sense, i.e. d²f/dx² ≥ 0. In particular, "concave-down" is NOT convex! §§ ** UPDATE: ** I think Cave has exactly the right approach; unfortunately, it suffers a bit in presentation, so I thought I'd dress it up a bit. [NOTE: Read carefully; it's very difficult in this font to distinguish f, f', and f'' -- that is, f, df/dx, and d²f/dx².] Integrating by parts, we have that ∫ f(x) cos(x) dx = f(x)·sinx - ∫ f'(x) sin(x) dx = - ∫ f'(x) sin(x) dx, since sin(x) = 0 at both endpoints. Now, since sin(x) = sin(π-x), we have that ∫ f(x) cos(x) dx = - ∫ f'(x) sin(x) dx = - ∫ [x=0,π] f'(x) sin(x) dx - ∫ [x=π,2π] f'(x) sin(π-x) dx = - ∫ [x=0,π] f'(x) sin(x) dx - ∫ [x=0,π] f'(x+π) sin(-x) dx = - ∫ [x=0,π] f'(x) sin(x) dx + ∫ [x=0,π] f'(x+π) sin(x) dx = ∫ [x=0,π] ( f'(x+π) - f'(x) )·sin(x) dx. Finally, since f is convex (i.e. concave-up), f' is increasing, and so f'(x+π) ≥ f'(x) for any x in [0,π]. Therefore we have that ( f'(x+π) - f'(x) )·sin(x) ≥ 0 on [0,π], and so the proof is complete. §§§
If f(x) =3x^2 and g(x) =x-1, how do you find f(g(x)) and g(f(x))?
These problems use the 'nested' function rule. If f(x) = 3x^2, and g(x) = x - 1:f(g(x)) means that the function g(x) itself will be used as an 'x' value in f(x)likewise, in g(f(x)) means that the function f(x) itself will be used as an 'x' value in g(x)Hence, f(g(x)) = f(x-1) = 3(x-1)^2.And g(f(x)) = g(3x^2) = (3x^2) - 1
Why is it that if f(g(x)) =g(f(x)) then f(x) and g(x) are inverses?
Important facts about inverses, where f(x) and g(x) are inverse functions:f(g(x)) = x g(f(x)) = xf(x) is the reflection of g(x) across the line y = xg(x) is the reflection of f(x) across the line y = xIf the point (a,b) is on the graph of f(x), then the point (b,a) is on the graph of g(x), and vice versa If you can't get it from that, it can always help to try specific functions that you KNOW are inverses. For example, you know that x^2 and SQRT(x) are inverses (as long as SQRT(x) is defined). But does f(x) = -g(x)? Not really. Does f(x) / g(x) = 1? Nope. Does f(g(3)) = 1? Or does it equal 3Free Calculus Help
University Calculus Integral Question!?
Let S(t) be a function such that S'(t) = sqrt(1+t^2). Let F(x) be a function such that F'(x) = f(x). From the fundamental theorem of calculus, we have f(x) = S(sin x) - S(0) g(y) = F(y) - F(3). Then, we have f'(x) = S'(sin x) * (d/dx of sin x) - 0 = (cos x) sqrt(1 + sin^2 x) from the chain rule g'(y) = F'(y) - 0 = f(y) g''(y) = f'(y) = (cos y) sqrt(1 + sin^2 y) g''(pi/6) = [cos (pi/6)] sqrt[1 + sin^2 (pi/6)] = [sqrt(3) / 2] sqrt(5/4) = sqrt(15) / 4. Lord bless you today!
How do you do this integral problem?
We use the Fundamental Theorem of Calculus with a twist: First the FTC states that as long as differentiability and continuity are fine, then if g(x) = ∫[a,x] f(t) dt, then g'(x) = f(x), or more simply we just have to plug in the x. Now the questions don't have a plain old x, the first one has a sin(x). Now remember when we differentiated other functions like g(x) = (x² - 1)³? We had to use chain rule. The same thing applies here. Now people like to use the substitution method of differentiating, so I will go that route: First, if we had the derivative of f(x) is df(x)/dx, we could say that this is the same as df(x)/du * du/dx, because the du's cancel out, so let's just keep that in mind: First of all, I'm assuming you want f '(x) for the first one: f(x) = ∫[0,sin(x)] √(1+t^2) dt Let u = sin(x), then du/dx = cos(x) Now, let's substitute and differentiate: f(x) = ∫[0,sin(x)] √(1+t^2) dt f(x) = ∫[0,u] √(1+t^2) dt df(x)/du = √(1+u^2) Now notice how there is a df(x)/du, and not df(x)/dx? That's because we changed variables, so we must change what we are differentiating with respect to. However, we still want to get to df(x)/dx. How do we do that? By using what we know from earlier: df(x)/dx = df(x)/du * du/dx df(x)/du = √(1+u^2), now just multiply both sides by du/dx df(x)/du * du/dx= √(1+u^2) * du/dx df(x)/dx = √(1+u^2) * du/dx, and now we substitute back what we know, which is u = sin(x) and du/dx = cos(x) df(x)/dx = √(1+u^2) * du/dx df(x)/dx = √(1+sin²(x)) * cos(x) This answer will come in handy for the actual evaluation: Now what is g''(y)? Well: g(y) = ∫[3,y] f(x) dx g'(y) = f(y) g''(y) = f '(y) g''(π/6) = f '(π/6) g''(π/6) = √(1+sin²(π/6)) * cos(π/6) g''(π/6) = √(1+(1/2)²) * (√3/2) g''(π/6) = √(5/4) * (√3/2) g''(π/6) = √5/2 * (√3/2) g''(π/6) = √15/4 g''(π/6) = 0.9682458365
If y = integral [0, g(x)] (1/sqrt(1+t^3)) dt, where g(x) = integral [0, cosx] (1+sin(t^2)) dt, find f ' (pi/2)?
Let : u = g(x). ∴ y = ∫ [0,u] [ 1 / √(1+t³) ] dt ... and ... u = ∫ [0,cos x] [ 1 + sin (t²) ] dt. ∴ dy/du = 1 / √(1+u³) ... and ... du/dx = [1 + sin ( cos² x )]· (-sin x)........ (1) ∴ by Chain Rule, ... ƒ'(x) = dy/dx = ( dy/du )·( du/dx ) ............ (2) ......................................... Now, at x = π/2, ... cos x = cos π/2 = 0 ∴ u = g(π/2) = ∫ [0,0] ... dt = 0 ∴ dy/du = 1 /√(1+0³) = 1.............................. (3) Also, ... du/dx = [ 1 + sin ( cos² π/2) ]· ( - sin π/2) = [ 1 + 0 ]·( -1 ) = -1.......... (4) ......................................... Finally, from (1) to (4), ƒ'(π/2) = (1)·(-1) = -1 ............ Ans. ......................................... Happy To Help ! .........................................