Maximizing Slope: Solving for Tangent Line on y=1+40x^3-3x^5 Curve

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Homework Statement



At which points on the curve y=1 + 40x^3 - 3x^5 does the tangent line have th largest slope?

Homework Equations



derivative is 120x^2-15x^4...

The Attempt at a Solution



how should i do this? start by setting the derivative equal to zero to find critical #s, but then what?
 
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It asks you to maximize the slope of the tangent line = to maximize the first derivative.
 

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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