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The Derivative Test is a key tool in calculus for determining the behavior of functions, such as identifying intervals where a function increases or decreases. In our exercise, we consider the function \( g(x) = e^x - (1 + x) \). To apply the Derivative Test, we first find the derivative \( g'(x) \). This involves differentiating each part separately:

• The derivative of \( e^x \) is \( e^x \).
• The derivative of \( 1 + x \) is \( 1 \).

Combining these, we get \( g'(x) = e^x - 1 \).Now, we analyze the sign of \( g'(x) \):

• For \( x \geq 0 \), \( e^x \geq 1 \), hence \( g'(x) \geq 0 \), indicating \( g(x) \) is non-decreasing.
• For \( x < 0 \), \( e^x \) approaches zero, but is always greater than \( 1 + x \), so \( g'(x) \) remains positive, meaning \( g(x) \) is still non-decreasing up to the inflection point.

Applying the First Derivative Test here shows how \( g(x) \) behaves and confirms that \( e^x \geq 1 + x \) for all \( x \). This test helps us conclude that the function never dips below zero, supporting the inequality.

Function Analysis is all about deeply understanding the behavior of a function through its structure and derivatives. In the exercise, we looked at the function \( g(x) = e^x - (1 + x) \) to show \( e^x \geq 1 + x \). Let's explore this further:By defining \( g(x) \), we put \( e^x \) and the linear function \( 1 + x \) into a single expression. Evaluating it at certain points provides key insights:

• At \( x = 0 \), \( g(0) = 0 \), meaning the functions intersect.
• The derivative, \( g'(x) = e^x - 1 \), shows the growth rate difference between \( e^x \) and the line \( 1+x \).

Next, we observe how changes in \( x \) affect \( g(x) \). For growing \( x \), \( e^x \)'s exponential increase ensures that \( g(x) \) consistently stays non-negative. For negative \( x \), the function \( e^x \) still dominates over \( 1 + x \) due to its slow decline compared to any linearly decreasing function. This use of function analysis through derivatives and their signs helps us understand the inequality without testing each possible \( x \).

Inequalities in Calculus often involve demonstrating that one function is greater or smaller than another over an interval. The exercise solves the inequality \( e^x \geq 1 + x \) using calculus.Consider how inequalities are substantiated:

• By constructing \( g(x) = e^x - (1 + x) \), we convert an inequality into an equality problem: \( g(x) \geq 0 \).
• The derivative \( g'(x) = e^x - 1 \) indicates where differences in slope may cause crossover points or direct comparison help confirm the inequality.

For \( x \geq 0 \), showing \( g(x) \geq 0 \) after evaluating at \( x = 0 \) and checking the derivative, it solidifies because the slope \( e^x - 1 \geq 0 \), meaning our gap widens or remains constant.Inequalities are visually intuitive too. An exponentially increasing \( e^x \) swiftly rises above the linear \( 1 + x \) even past zero, showing a clear dominance. Furthermore, calculus ensures this is true across any real number \( x \) without extensive numerical calculation. This exercise highlights how calculus and simple derivative checks can effectively uphold inequalities.