The derivative \( f'(x) = rac{-5}{(x - 1)^2} \) is always negative for all \( x
["Understanding Why the Derivative ( f'(x) = -\frac{5}{(x - 1)^2} ) Is Always Negative", "In calculus, one of the most basic yet essential tasks is analyzing the sign of a derivative to determine the behavior of a function—specifically, whether it is increasing or decreasing on a given interval. For the derivative given by\n[\nf'(x) = -\frac{5}{(x - 1)^2},\n]\na common question is: why is this derivative always negative for all ( x )? This article explains the reasoning step-by-step and explores the mathematical implications.", "---", "### What Does It Mean for ( f'(x) ) to Be Negative?", "The sign of the derivative ( f'(x) ) tells us about the slope of the original function:", "- If ( f'(x) > 0 ) for all ( x ) in an interval, ( f(x) ) is increasing on that interval.\n- If ( f'(x) < 0 ) for all ( x ) in an interval, ( f(x) ) is decreasing on that interval.", "Thus, knowing that ( f'(x) = -\frac{5}{(x - 1)^2} ) is always negative helps us immediately conclude that the function ( f(x) ) is strictly decreasing everywhere except at ( x = 1 ), where the derivative is undefined.", "---", "### Analyzing the Expression", "Let’s break down the derivative:\n[\nf'(x) = -\frac{5}{(x - 1)^2}\n]", "1. Numerator: The constant (-5) is negative.\n2. Denominator: The term ( (x - 1)^2 ) is a square, which is always non-negative (i.e., ( (x - 1)^2 \geq 0 )) for all real ( x ).\n - Importantly, it is never zero except when ( x = 1 ), where the denominator is zero, making ( f'(x) ) undefined.", "Since a negative number divided by a positive number is negative, and the denominator is either positive or undefined (never zero in the domain of ( f'(x) )), we conclude:", "- For all ( x <br/>\neq 1 ), ( f'(x) < 0 ).\n- At ( x = 1 ), the derivative does not exist due to division by zero, but we can define ( f(x) ) to be decreasing on both sides.", "---", "### Graphical Interpretation", "The sign of the derivative directly affects the graph:\n- Since ( f'(x) < 0 ) for all ( x <br/>\ne 1 ), the function ( f(x) ) is always decreasing on its domain ( (-\infty, 1) \cup (1, \infty) ).\n- This explains why the graph slopes downward everywhere except at ( x = 1 ), where there may be a vertical asymptote or removable discontinuity depending on the full function, but no local increase.", "---", "### Why is the Derivative Never Zero?", "For ( f'(x) = -\frac{5}{(x - 1)^2} ) to be zero, the numerator would need to be zero. However, the numerator is (-5), which is never zero. Even though the denominator vanishes at ( x = 1 ), producing an undefined (infinite) slope, the derivative itself never crosses zero—it remains strictly negative.", "---", "### Practical Implications", "Knowing that ( f'(x) < 0 ) for all ( x ) allows us to:", "- Confirm that the function has no local maxima or minima (except possibly at discontinuities).\n- Apply the First Derivative Test confidently to determine monotonicity.\n- Reason about integrals and areas under the curve with consistent sign behavior.\n- Understand behavior of related functions and model real-world processes like decay rates.", "---", "### Final Summary", "The derivative\n[\nf'(x) = -\frac{5}{(x - 1)^2}\n]\nis always negative for all real ( x ) except ( x = 1 ), where it is undefined. The negative sign arises from the (-5) in the numerator combined with the always-positive square denominator. This confirms that the original function is strictly decreasing everywhere in its domain, providing a clean and definitive behavior pattern useful in calculus and applied mathematics.", "---", "Keywords: derivative is always negative, negative derivative, f'(x), calculus, function behavior, decreasing function, analyzing the sign of the derivative, x-interval behavior, mathematical reasoning.\nMeta Description: Discover why ( f'(x) = -\frac{5}{(x - 1)^2} ) is always negative — learn how the sign of a derivative determines function monotonicity in calculus."]
