F(x) = x^4 + ax^2 +b. the graph of x has a relative max at 0,1 and

meredith
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f(x) = x^4 + ax^2 +b. the graph of x has a relative max at 0,1 and an inflection point at x=1. the values of a and b are...?

ok I am really stuck.
thanks in advance!
 
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you have a point on the graph (0,1), when x=0, f(x)=1...you can get one constant.

at a point of inflexion f''(x)=0
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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