The greatest and lower value of the function in this segment.

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What is the greatest and lower value of the functiony=x+\sqrt{x}
in the segment [0,4]?
I knowit has to do with deriving it , but i don't know how to explain it.
Any help :D
 
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The maximum and minimum of a continuous function f on [a,b] can only be in at a, b or a number c which satisfies f'(c)=0.
 
Thanks that's gives a clue.
So i replace a and b in the function , then equals it to 0 and compare a ,b and c. Isnt it so?
 
yes that's correct.
 
Penultimate said:
Thanks that's gives a clue.
So i replace a and b in the function , then equals it to 0 and compare a ,b and c. Isnt it so?
This is pretty vague, so you might understand what you mean, but you're not stating it clearly. What you need to do is to evaluate f(0) and f(4). Then you need to find f'(x) and set it to zero. If there is a number c for which f'(c) = 0, then evaluate f(c). Of the three numbers f(0), f(4), and f(c), one of them will be the largest function value on the interval [0, 4] and one will be the smallest.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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