Question

Let $f(x) = \frac{\log(e + x)}{\log(\pi + x)}, \; -2 < x < \infty$. Then $f$ is

• A. decreasing on (-2,$\infty$)
• B. decreasing only on (0,$\infty$)
• C. increasing only on (0, $\pi$)
• D. increasing on (-2,$\infty$)
• E. increasing only on (0,$\pi$)

Hint:

Math Tip: To quickly determine if an expression like $A\ln A - B\ln B$ is positive, observe that the function $x\ln x$ is increasing for $x>1/e$. If $A>B>1/e$, then $A\ln A$ will always be greater than $B\ln B$.

Answer 1

The Correct Option is D

Concept:
Calculus - Monotonicity and Application of Derivatives.
A function $f(x)$ is increasing on an interval if its first derivative $f'(x)>0$ for all $x$ in that interval. Note: "$\log$" in calculus standard notation implies the natural logarithm "$\ln$".
Step 1: Differentiate the function using the quotient rule.
Given $f(x) = \frac{\ln(e+x)}{\ln(\pi+x)}$. $$ f'(x) = \frac{\frac{d}{dx}[\ln(e+x)] \cdot \ln(\pi+x) - \ln(e+x) \cdot \frac{d}{dx}[\ln(\pi+x)]}{[\ln(\pi+x)]^2} $$ $$ f'(x) = \frac{\left(\frac{1}{e+x}\right)\ln(\pi+x) - \left(\frac{1}{\pi+x}\right)\ln(e+x)}{[\ln(\pi+x)]^2} $$
Step 2: Simplify the numerator.
Find a common denominator for the terms in the numerator: $$ \text{Numerator} = \frac{(\pi+x)\ln(\pi+x) - (e+x)\ln(e+x)}{(e+x)(\pi+x)} $$
Step 3: Analyze the components to determine the sign of $f'(x)$.
Let's define a helper function $g(t) = t\ln t$. The numerator of our derivative contains the expression $g(\pi+x) - g(e+x)$. We must determine if this expression is positive or negative.
Step 4: Check the monotonicity of the helper function $g(t)$.
Find the derivative of $g(t)$: $$ g'(t) = (1)\ln t + t\left(\frac{1}{t}\right) = \ln t + 1 $$ For $g'(t)>0$, we need $\ln t + 1>0 \implies \ln t>-1 \implies t>e^{-1} \approx 0.36$. Thus, $g(t)$ is strictly increasing for all $t>0.36$.
Step 5: Apply the domain restrictions.
The given domain is $x>-2$. Let's evaluate our inputs to $g(t)$:

• $e+x>e - 2 \approx 2.718 - 2 = 0.718$
• $\pi+x>\pi - 2 \approx 3.141 - 2 = 1.141$

Since both inputs are greater than $0.36$, they fall in the strictly increasing region of $g(t)$.
Step 6: Determine the final sign of the derivative.
Because $\pi>e$, it is always true that $(\pi+x)>(e+x)$. Since $g(t)$ is an increasing function, larger inputs yield larger outputs: $$ g(\pi+x)>g(e+x) \implies g(\pi+x) - g(e+x)>0 $$ Therefore, the numerator is strictly positive. The denominator $(e+x)(\pi+x)[\ln(\pi+x)]^2$ is also positive. Since $f'(x)>0$ for all $x>-2$, the function is increasing on $(-2, \infty)$.