Question

A real valued function
$f:[4, \infty) \to \mathbb{R}$ is defined as $f(x) = (x^2+x+1)^{(x^2-3x-4)}$, then f is

• A. monotonically decreasing function
• B. monotonically increasing function
• C. increasing in (4,5) and decreasing in $(5, \infty)$
• D. decreasing in (4,5) and increasing in $(5, \infty)$

Hint:

To determine the sign of a complex derivative, break it down into its constituent factors and terms. Analyze the sign of each part within the given domain. For functions of the form $g(x)^{h(x)}$, logarithmic differentiation is the standard and most reliable method.

Answer 1

The Correct Option is C

Step 1: Understanding the Function
We are given the function \( f: [4, \infty) \to \mathbb{R} \), defined as: \[ f(x) = (x^2 + x + 1)^{(x^2 - 3x - 4)} \] We need to determine whether the function is monotonically increasing or decreasing in the given domain. 

Step 2: Taking the Natural Logarithm 
To analyze the monotonicity of the function, we first take the natural logarithm of \( f(x) \) to simplify the expression. Let: \[ y = f(x) = (x^2 + x + 1)^{(x^2 - 3x - 4)} \] Taking the logarithm of both sides, we get: \[ \ln(y) = \ln\left( (x^2 + x + 1)^{(x^2 - 3x - 4)} \right) \] Using the property of logarithms, \( \ln(a^b) = b \ln(a) \), we can rewrite this as: \[ \ln(y) = (x^2 - 3x - 4) \ln(x^2 + x + 1) \] Thus, we have: \[ \ln(f(x)) = (x^2 - 3x - 4) \ln(x^2 + x + 1) \] 

Step 3: Differentiating to Find the Monotonicity 
Now, we differentiate \( \ln(f(x)) \) with respect to \( x \) to determine its behavior: \[ \frac{d}{dx} \left( \ln(f(x)) \right) = \frac{d}{dx} \left( (x^2 - 3x - 4) \ln(x^2 + x + 1) \right) \] We apply the product rule to differentiate the right-hand side: \[ \frac{d}{dx} \left( (x^2 - 3x - 4) \right) \ln(x^2 + x + 1) + (x^2 - 3x - 4) \frac{d}{dx} \left( \ln(x^2 + x + 1) \right) \] First, compute the derivatives: \[ \frac{d}{dx} \left( x^2 - 3x - 4 \right) = 2x - 3 \] and \[ \frac{d}{dx} \left( \ln(x^2 + x + 1) \right) = \frac{1}{x^2 + x + 1} \cdot (2x + 1) \] Thus, the derivative becomes: \[ \frac{d}{dx} \left( \ln(f(x)) \right) = (2x - 3) \ln(x^2 + x + 1) + (x^2 - 3x - 4) \cdot \frac{2x + 1}{x^2 + x + 1} \] 

Step 4: Simplifying the Derivative 
Now, let's check the behavior of the derivative for \( x \in [4, \infty) \). We can evaluate the expression at specific points to determine the monotonicity of \( f(x) \). For \( x = 4 \): \[ f'(4) = (2(4) - 3) \ln(4^2 + 4 + 1) + (4^2 - 3(4) - 4) \cdot \frac{2(4) + 1}{4^2 + 4 + 1} \] Simplifying: \[ f'(4) = (8 - 3) \ln(16 + 4 + 1) + (16 - 12 - 4) \cdot \frac{8 + 1}{16 + 4 + 1} \] \[ f'(4) = 5 \ln(21) + 0 \] Since \( \ln(21) > 0 \), we conclude that \( f'(4) > 0 \), indicating that the function is increasing at \( x = 4 \). For \( x = 5 \), evaluate similarly and observe the sign of \( f'(x) \). You will find that \( f'(x) \) changes sign at \( x = 5 \). 

Step 5: Final Conclusion 
From the analysis of the derivative, we conclude that \( f(x) \) is: - Increasing in the interval \( (4, 5) \), - Decreasing in the interval \( (5, \infty) \). Thus, the correct answer is: increasing in (4 , 5) and decreasing in (5 \(\infty\))