Problem 27 - Essential Calculus Early Transcendentals

Question

Show that \(\sqrt{1+x}<1+\frac{1}{2} x\) if \(x>0\)

Step-by-Step Solution

Answer

The inequality holds true for all \( x > 0 \).

Step 1: Assume the Expression

To show the inequality \(\sqrt{1+x}<1+\frac{1}{2} x\), we start by considering \( x > 0 \). For simplification, assume \( y = \sqrt{1+x} \), so \( y^2 = 1 + x \).

Step 2: Rearrange the Inequality

The given inequality can be rewritten as \( \sqrt{1+x} - 1 < \frac{1}{2}x \). Using \( y^2 = 1 + x \), substitute \( x = y^2 - 1 \) into the inequality: \[ y - 1 < \frac{1}{2}(y^2 - 1) \].

Step 3: Clear the Fraction

Multiply the entire inequality by 2 to clear the fraction:\[ 2(y - 1) < y^2 - 1 \].This simplifies to:\[ 2y - 2 < y^2 - 1 \].

Step 4: Rearrange Terms

Rearrange the terms in the inequality to have all terms on one side:\[ y^2 - 2y + 1 > 0 \].

Step 5: Recognize the Perfect Square

Notice that the expression on the left is a perfect square:\[ (y - 1)^2 > 0 \]. Since \( y = \sqrt{1+x} \), and \( x > 0 \), then \( y = \sqrt{1+x} > 1 \), implying \( y eq 1 \). This confirms that \( (y - 1)^2 > 0 \).

Step 6: Conclusion

The inequality \( (y - 1)^2 > 0 \) is satisfied for all \( x > 0 \) since \( y \) cannot be 1. Therefore, the original inequality \( \sqrt{1+x} < 1 + \frac{1}{2}x \) holds true for all \( x > 0 \).

Key Concepts

Proofs in Calculus

A proof in calculus is a logical argument that demonstrates why a particular statement or theorem is true. Proofs often involve algebraic manipulations, substitutions, and reasoning to transition from assumptions to conclusions. In our exercise, the key is showing that the inequality \( \sqrt{1+x} < 1 + \frac{1}{2}x \) holds for \( x > 0 \).

Proofs can take different forms, such as direct proofs, indirect proofs like contradiction, or proofs by induction. In this case, it's a direct approach. We begin by assuming the inequality, then manipulate it and apply some algebraic principles to verify it. The beauty of calculus proofs lies in their precision. Each step follows logically from the last, resulting in a sound conclusion. Proofs help us deepen our understanding of the foundational concepts in calculus.

Square Root Function

The square root function is an essential part of calculus, turning inputs into their square roots. The function \( f(x) = \sqrt{x} \) only has outputs for non-negative values of \( x \). It is continuous and increasing for \( x \geq 0 \).

If we consider \( \sqrt{1+x} \), it shows how square root functions can be manipulated within inequalities. This function becomes vital when dealing with expressions involving squares, as it simplifies square expressions.

In the example, the property \( y = \sqrt{1+x} \) is at the center of the proof. We recognize that if \( x > 0 \), then \( 1 + x > 1 \), meaning the square root is greater than 1 but not equal to 1. This fact becomes crucial in manipulating the inequality trustfully.

Perfect Square

Perfect squares are expressions like \( a^2 \), meaning they are the result of squaring a number or expression. They can simplify calculations and are useful in confirming inequalities. They help by giving exact values for comparisons.

In this exercise, the expression \( y^2 - 2y + 1 \) is a perfect square rewritten as \( (y - 1)^2 \). Recognizing perfect squares is a handy skill. It enables a better understanding of the relationships between algebraic expressions.

The identification of \( (y - 1)^2 \) shows us that for any non-zero \( y = \sqrt{1+x} \), the square of the difference from 1 will always be greater than zero. This conclusion is used to justify the inequality in our proof, since it parallels the form \( y^2 - 2y + 1 > 0 \).

Inequality Manipulation

Manipulating inequalities is about adjusting and transforming inequalities while keeping their truth. It's a skill central to calculus, often involving addition, subtraction, multiplication, division, and factoring.

In this proof, the main goal was to manipulate \( \sqrt{1+x} < 1 + \frac{1}{2}x \) into a form that's easy to analyze. By multiplying and rearranging terms, we transformed the inequality into \( (y - 1)^2 > 0 \). Each of these steps follows rules that ensure the inequality's direction (\(<\) vs. \(>\)) stays correct.

Understanding how changes affect inequality is key to achieving correct solutions. Specific transformations, like clearing fractions by multiplication, help simplify expressions, making the logic easier to follow. Practicing these techniques bridges algebra with calculus, refining problem-solving skills.