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**Tweety** Find the coordinates of the turning point on the graph of $\displaystyle y = x\times e^{2x} $ and determine its

nature.

I used the product rule and got $\displaystyle e^{2x} +2x\times e^{2x} $

so to find the coordinates of the turning point I set that first derivative equal to 0.

$\displaystyle e^{2x} +2x\times e^{2x} = 0 $

$\displaystyle e^{2x} (1+2x) = 0 $

$\displaystyle e^{2x} = 0 $ or $\displaystyle 1+2x = 0 $

$\displaystyle x = -\frac{1}{2} $

and to find the 'nature' of the turning points I have to take the second derivative. in which I dont get the correct answer, heres my working,

$\displaystyle \frac{dy}{dx} = e^{2x} + 2xe^{2x} $

the derivative of $\displaystyle e^{2x} = 2e^{2x} $

using the product rule on $\displaystyle 2x e^{2x} $

$\displaystyle 2 \times e^{2x} + 2e^{2x} \times 2x $

$\displaystyle \frac{d^{2}y}{dx^{2}} = 2e^{2x} + 2 \times e^{2x} + 2e^{2x} \times 2x $

= $\displaystyle 4e^{2x} (1+x) $

but the my book says that the correct answer is $\displaystyle \frac{d^{2}y}{dx^{2}} = 2(2x+1)e^{2x} + e^{2x}.2 $