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: Analyze the base and exponent Let \( u(x) = x^2 + x + 1 \) and \( v(x) = x^2 - 3x - 4 \). Then \( f(x) = u(x)^{v(x)} \). For \( x \in [4, \infty) \): Base \( u(x) = x^2 + x + 1 \). Since \( x \ge 4 \), \( u(x)>1 \) and \( u'(x) = 2x + 1>0 \). So the base is increasing. Exponent \( v(x) = x^2 - 3x - 4 = (x-4)(x+1) \). For \( x>4 \), \( v(x)>0 \) and \( v'(x) = 2x - 3>0 \) (since \( 2(4)-3=5>0 \)). So the exponent is positive and increasing.
Step 2: Determine monotonicity Since both the base \( u(x)>1 \) is increasing and the exponent \( v(x)>0 \) is increasing for \( x \ge 4 \), the function \( f(x) \) must be strictly increasing. Mathematically, \( \ln f(x) = v(x) \ln u(x) \). \( \frac{f'(x)}{f(x)} = v'(x) \ln u(x) + v(x) \frac{u'(x)}{u(x)} \). Since \( x \ge 4 \): \( v'(x)>0 \), \( \ln u(x)>0 \) (as \( u(x)>1 \)), \( v(x) \ge 0 \), \( u'(x)>0 \), \( u(x)>0 \). Thus, \( f'(x)>0 \) for \( x>4 \).