The function f(x) = e^3x +6x^2 +1 has a horizontal tangent at x =?

meredith
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Homework Statement




i can't solve it!
im really lost. i know you find f''(x) (i got 9e^3x +12) but i don't know where to go from there
what would i do on my calculator?
 
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to calculate the horizontal tangent you only need to solve for f ' (x) = 0. f '' (x) is used and further derivatives are used to check whether that particular solution of the first equation gives a maxima, minima or an inflexion point.

Solve 3e^(3x) + 12x = 0
 
And that equation you will need to solve numerically, perhaps by graphing it on your graphing calculator and "zooming" in where it crosses the x-axis.
 

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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