Max/Min of Polynomial Function

Hollysmoke
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The question is:

Determine the equation of the line tangent to f(x)=4x^3+12x^2-96x wit hthe smallest slope on the interval -4 <= x <= 2.

So I found the derivative and critical numbers to find the max and min but where do I go from there?
 
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So you need to find the line with minimum slope. What function gives you the slope? How do you find extremes of a function?
 
By finding the critical numbers of the derivative and plugging them into the original function, as well as the intervals and finding the lowest number would be the minimum value.
 
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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