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    Hi guys, stuck on a C3 question...

    The question asks me to prove f(x) = sinx-xcosx through the interval 0 to pi/2

    I've differentiated it to give xsin(x), but I'm not sure where to go from here.
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    (Original post by decb24)
    Hi guys, stuck on a C3 question...

    The question asks me to prove f(x) = sinx-xcosx through the interval 0 to pi/2

    I've differentiated it to give xsin(x), but I'm not sure where to go from here.
    What was your aim when you decided to differentiate f(x)? Use this to help you; and also pay attention to the interval given.
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    If f(x) is increasing over that range then f'(x) will be positive over that range. We know that x is not negative between 0 and pi/2. How about sine? What happens when you multiply two positive numbers together?
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    Perhaps draw a graph of sinx as a reminder to you.

    Now you want to consider f'(x) over [0,pi/2], so what is required of f'(x) for f(x) to be an increasing function.

    Remember f'(x) represents the gradient of the function
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    So f'(x) needs to be greater than 0, and that could be proved with an inequality, but I'm not sure how the interval comes into this.
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    (Original post by decb24)
    So f'(x) needs to be greater than 0, and that could be proved with an inequality, but I'm not sure how the interval comes into this.
    well, is xSinx always positive?
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    (Original post by TenOfThem)
    well, is xSinx always positive?

    No, but when solving the inequality how do I factor in the fact that I'm only proving this within an interval?
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    (Original post by decb24)
    No, but when solving the inequality how do I factor in the fact that I'm only proving this within an interval?
    You say

    0<x<pi/2 means that ... ... ...
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    for f ' (x)&gt;0 we already know that x&gt;0

    so that just leaves \sin(x)&gt; to make (+ve) x (+ve) = +ve => increasing

    is Sine positive in [0,Pi/2] ? (you can prove this with a graph, or by using the identity:

    \sin(x)= +\sqrt{1-\cos^{2}(x)}) - which will be less than 1, but....?) - (because you are in the "ALL" quadrant of the CAST diagram)

    EDIT: but, to be honest - TenOfThem has it...
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    Slightly off topic, but a function increasing when the derivative is >= 0 or just >0, and why is this? I'm sure I've seen different versions of this in different questions. :holmes:
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    >0

    =0 => stationary point.
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    (Original post by Hasufel)
    >0

    =0 => stationary point.
    For example, if we take the function f(x)=x^2, isn't it increasing when x <= 0?

    The definitions I've seen of increasing and strictly increasing seem to suggest that the point x=0 would also be included, but I might just be getting mixed up

    Posted from TSR Mobile
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    (Original post by Arithmeticae)
    For example, if we take the function f(x)=x^2, isn't it increasing when x <= 0?
    No, because f(-1) < f(-2), although I suspect this is just a typo and you mean "when x \ge 0.

    But even then, at x = 0, we find y arbitrarily close to x with y < x but f(y) > f(x). (Because f(x) is decreasing for x < 0). So f isn't increasing at 0.
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    (Original post by Arithmeticae)
    Slightly off topic, but a function increasing when the derivative is >= 0 or just >0, and why is this? I'm sure I've seen different versions of this in different questions. :holmes:
    If f is defined and continuous on the interval [a,b], and f is differentiable on the interval (a,b), then applying the Mean Value Theorem tells us if a \leq x &lt; y \leq b, then f(y) = f(x) + (y-x) f'(\xi) for some \xi between x and y. If we know f'(\xi) &gt; 0 whenever a < x < b,then we must have f(y) > f(x) and so f is increasing. If we know f'(\xi) \geq 0 whenever a < x < b,then we must have f(y) \geq f(x) and so f is non-decreasing.

    Note: I've made the distinction between "increasing" and "non-decreasing" here. "increasing" is slightly ambiguous, as sometimes people use it when they mean x &lt; y \implies f(x) &lt; f(y) (1) and sometimes when they mean x &lt; y \implies f(x) \leq f(y) (2). The least ambiguous terms are "strictly increasing" for case (1), and non-decreasing for case (2).

    As far as derivatives go, if f'(x) = 0, it requires more complex tests to tell whether f is increasing or decreasing. You can look at higher derivatives: if the kth deriviative is the first non-zero derivative, then if k is odd, f is increasing if the kth derivative is positive, decreasing otherwise. If k is even, f is a max/min point, so increasing on one side of x and decreasing on the other,

    But, there are non constant functions where you can find points such that all the deriviatives at that point are zero. At which point you just have to look at how the function actually behaves.

    e.g. f(x) = \exp(-1/x^2) (x \neq 0), f(0) = 0 can be shown to be infinitely differentiable at 0, with f^{(k)}(0) = 0 for all k. But looking at the function, it's clear 0 is a minimum point.

    Or f(x) = -\exp(-1/x^2) (x &lt; 0), exp(-1/x^2) (x &gt; 0), f(0) = 0 is again infinitely differentiable at 0 with f^{(k)}(0) = 0 for all k. But looking at the function, it's clear it's increasing near 0.
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    (Original post by Arithmeticae)
    For example, if we take the function f(x)=x^2, isn't it increasing when x <= 0?

    The definitions I've seen of increasing and strictly increasing seem to suggest that the point x=0 would also be included, but I might just be getting mixed up

    Posted from TSR Mobile
    I've been taught that increasing means that it doesn't get smaller (so weak inequalities).
 
 
 
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