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Find the coordinates of stationary points on the curve with the equation

$(y-2)^2e^x=4x$

I differentiated to get

$2(y-2)e^x\frac{dy}{dx}+(y-2)^2e^x=4$

$\frac{dy}{dx}=0$

$(y^2-4y+4)e^x=4$

What do I do from here?

Jim
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2 Answers2

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Since at the stationary point we have $\frac{dy}{dx}=0$, set it equal to $0$ in your equation $2(y-2)e^x\frac{dy}{dx}+(y-2)^2e^x=4$.

We get that $(y-2)^2 e^x=4$. Comparing with the equation $(y-2)^2e^x=4x$, we get $x=1$.

André Nicolas
  • 507,029
  • I ended up with $\frac{2e±√e}{e}$. The answer in the back is $y=\frac{2\sqrt e \pm 2}{\sqrt e}$ Where am I going wrong? – Jim May 16 '14 at 19:22
  • We get $(y-2)^2=\frac{4}{e}$, so $y=2\pm\frac{2}{\sqrt{e}}$. This can be transformed to the answer in the back, or to $\frac{2e+\pm 2\sqrt{e}}{e}$, but not to your expression. But the second is close to your expression, probably a small algebra glitch. – André Nicolas May 16 '14 at 19:33
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You're almost there. Substitute curve equation $(y-2)^2 e^x = 4x$ to your last equation, so you can get $4x = 4$ or equivalently $x = 1$. Now, solve for $y$'s $$ (y-2)^2e=4 \implies y = 2 \pm \frac 2{\sqrt e} = \frac {2\sqrt e \pm 2}{\sqrt e} = \frac {2e \pm 2\sqrt e}e $$

Here's some primitive Mathematica code to see if everything is right

sol = Solve[(y - 2)^2 E == 4, y];
x = 1;
y1 = y /. sol[[1]];
y2 = y /. sol[[2]];
plot = ContourPlot[(y - 2)^2 E^x - 4 x == 0, {x, 0, 8}, {y, 0.5, 3.5},
    AspectRatio -> Automatic];
top = Graphics[{Circle[{x, y2}, 0.05]}];
bottom = Graphics[{Circle[{x, y1}, 0.05]}];
topT = Graphics[{Red, Line[{{0, y2}, {8, y2}}]}];
bottomT = Graphics[{Red, Line[{{0, y1}, {8, y1}}]}];
Show[plot, top, bottom, topT, bottomT]

plot

Kaster
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