Find Maximum Value of f(x)+x | Derivative Question

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1. f(x)=-x2+4x-5 (parabola)

What's the maximum value of "f(x)+x"?


sorry for my english :)
 
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Clu2.0 said:
1. f(x)=-x2+4x-5 (parabola)

What's the maximum value of "f(x)+x"?


sorry for my english :)
Since f(x)= -x2+ 4x- 5, f(x)+ x= -x2+ 5x- 5. That can be written as -(x2- 5x- 5. Now complete the square to dotermine where the vertex is.
 
thanks ok find the answer ("1") but in the test answer is 5/4 and teachers says it's correct
 
Isn't this just a simple case of taking the first derivative with respect to x and setting it equal to zero? This then tells you the value of x where the maximum value occurs. Then substituting this value into the function f(x)+x will tell you the maximum value at that point. This procedure does give the correct answer of 5/4.
 
thanks a lot, my carelessness
 
Or via completing the square.
 
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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