Question

Consider $f(x)=x^{\ln x}$ and $g(x)=e^2x$. Let $\alpha$ and $\beta$ be two values of $x$ satisfying $f(x)=g(x)$ ($\alpha < \beta$)

If $h(x)=\frac{f(x)}{g(x)}$ then $h'(\alpha)$ equals to:

• A. e
• B. -e
• C. 3e
• D. -3e

Answer 1

Answer: (D)

-3e

Answer 2

To solve the problem, we follow these steps:

1. Simplify $f(x)$ and set up the equation $f(x)=g(x)$: Given $f(x) = x^{\ln x}$. We can rewrite this using the property $a^b = e^{b \ln a}$: $f(x) = e^{\ln(x^{\ln x})} = e^{(\ln x)(\ln x)} = e^{(\ln x)^2}$.

Given $g(x) = e^2 x$. Now, set $f(x) = g(x)$: $e^{(\ln x)^2} = e^2 x$

2. Solve for $x$ to find $\alpha$ and $\beta$: Take the natural logarithm on both sides of the equation: $\ln(e^{(\ln x)^2}) = \ln(e^2 x)$ $(\ln x)^2 = \ln(e^2) + \ln x$ $(\ln x)^2 = 2 + \ln x$

Rearrange into a quadratic equation by letting $y = \ln x$: $y^2 - y - 2 = 0$

Factor the quadratic equation: $(y-2)(y+1) = 0$

This gives two possible values for $y$: $y = 2 \quad \Rightarrow \ln x = 2 \quad \Rightarrow x = e^2$ $y = -1 \quad \Rightarrow \ln x = -1 \quad \Rightarrow x = e^{-1} = \frac{1}{e}$

We are given that $\alpha$ and $\beta$ are two values of $x$ satisfying $f(x)=g(x)$ and $\alpha < \beta$. Comparing the values $e^2$ and $\frac{1}{e}$: Since $e \approx 2.718$, $e^2 \approx 7.389$ and $\frac{1}{e} \approx 0.368$. Therefore, $\alpha = \frac{1}{e}$ and $\beta = e^2$.

3. Define and simplify $h(x)$: Given $h(x) = \frac{f(x)}{g(x)}$. Substitute the simplified forms of $f(x)$ and $g(x)$: $h(x) = \frac{e^{(\ln x)^2}}{e^2 x}$ Using properties of exponents ($e^a/e^b = e^{a-b}$) and $x^{-1} = e^{-\ln x}$: $h(x) = e^{(\ln x)^2 - 2} \cdot x^{-1}$ $h(x) = e^{(\ln x)^2 - 2} \cdot e^{-\ln x}$ $h(x) = e^{(\ln x)^2 - \ln x - 2}$

4. Differentiate $h(x)$: Use the chain rule for $h(x) = e^{u(x)}$ where $u(x) = (\ln x)^2 - \ln x - 2$: $h'(x) = e^{u(x)} \cdot u'(x)$ $u'(x) = \frac{d}{dx} ((\ln x)^2 - \ln x - 2)$ $u'(x) = 2 \ln x \cdot \frac{1}{x} - \frac{1}{x} - 0$ $u'(x) = \frac{2 \ln x - 1}{x}$

So, $h'(x) = e^{(\ln x)^2 - \ln x - 2} \cdot \frac{2 \ln x - 1}{x}$.

5. Evaluate $h'(\alpha)$: Substitute $\alpha = \frac{1}{e}$ into $h'(x)$. First, find $\ln \alpha$: $\ln \alpha = \ln\left(\frac{1}{e}\right) = \ln(e^{-1}) = -1$.

Now, substitute $\alpha = \frac{1}{e}$ and $\ln \alpha = -1$ into the expression for $h'(\alpha)$: The exponent in $e^{(\ln \alpha)^2 - \ln \alpha - 2}$ becomes: $(\ln \alpha)^2 - \ln \alpha - 2 = (-1)^2 - (-1) - 2 = 1 + 1 - 2 = 0$. So, $e^{(\ln \alpha)^2 - \ln \alpha - 2} = e^0 = 1$.

Substitute this back into $h'(\alpha)$: $h'(\alpha) = 1 \cdot \frac{2(\ln \alpha) - 1}{\alpha}$ $h'(\alpha) = 1 \cdot \frac{2(-1) - 1}{1/e}$ $h'(\alpha) = \frac{-2 - 1}{1/e}$ $h'(\alpha) = \frac{-3}{1/e}$ $h'(\alpha) = -3e$

The final answer is -3e.