QuestionJuly 14, 2025

The graph of y=(2x^2)^x has two horizontal tangent lines. Find equations for both of them.

The graph of y=(2x^2)^x has two horizontal tangent lines. Find equations for both of them.
The graph of y=(2x^2)^x has two horizontal tangent lines. Find equations for both of them.

Solution
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Answer

y = e^{-1} and y = e^{-1} (same equation for both horizontal tangents). Explanation 1. Find the derivative Let y = (2x^2)^x. Rewrite as y = e^{x \ln(2x^2)}. Differentiate using chain rule: \frac{dy}{dx} = y \cdot \frac{d}{dx}(x \ln(2x^2)). 2. Differentiate the exponent \frac{d}{dx}(x \ln(2x^2)) = \ln(2x^2) + x \cdot \frac{d}{dx}(\ln(2x^2)) = \ln(2x^2) + x \cdot \frac{2}{2x^2} \cdot 4x = \ln(2x^2) + 2. 3. Set derivative to zero for horizontal tangents \frac{dy}{dx} = y (\ln(2x^2) + 2) = 0. Thus, \ln(2x^2) + 2 = 0. Solve for x: \ln(2x^2) = -2. 4. Solve for x 2x^2 = e^{-2}, so x^2 = \frac{e^{-2}}{2}. Therefore, x = \pm \sqrt{\frac{e^{-2}}{2}}. 5. Find y values for each x For x = \sqrt{\frac{e^{-2}}{2}}, y = (2(\sqrt{\frac{e^{-2}}{2}})^2)^{\sqrt{\frac{e^{-2}}{2}}} = e^{-1}. For x = -\sqrt{\frac{e^{-2}}{2}}, y = (2(-\sqrt{\frac{e^{-2}}{2}})^2)^{-\sqrt{\frac{e^{-2}}{2}}} = e^{-1}.

Explanation

1. Find the derivative<br /> Let $y = (2x^2)^x$. Rewrite as $y = e^{x \ln(2x^2)}$. Differentiate using chain rule: $\frac{dy}{dx} = y \cdot \frac{d}{dx}(x \ln(2x^2))$.<br /><br />2. Differentiate the exponent<br /> $\frac{d}{dx}(x \ln(2x^2)) = \ln(2x^2) + x \cdot \frac{d}{dx}(\ln(2x^2)) = \ln(2x^2) + x \cdot \frac{2}{2x^2} \cdot 4x = \ln(2x^2) + 2$.<br /><br />3. Set derivative to zero for horizontal tangents<br /> $\frac{dy}{dx} = y (\ln(2x^2) + 2) = 0$. Thus, $\ln(2x^2) + 2 = 0$. Solve for $x$: $\ln(2x^2) = -2$.<br /><br />4. Solve for $x$<br /> $2x^2 = e^{-2}$, so $x^2 = \frac{e^{-2}}{2}$. Therefore, $x = \pm \sqrt{\frac{e^{-2}}{2}}$.<br /><br />5. Find $y$ values for each $x$<br /> For $x = \sqrt{\frac{e^{-2}}{2}}$, $y = (2(\sqrt{\frac{e^{-2}}{2}})^2)^{\sqrt{\frac{e^{-2}}{2}}} = e^{-1}$.<br /> For $x = -\sqrt{\frac{e^{-2}}{2}}$, $y = (2(-\sqrt{\frac{e^{-2}}{2}})^2)^{-\sqrt{\frac{e^{-2}}{2}}} = e^{-1}$.
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