Problem 57 - Calculus with Concepts in Calculus

Question

Show that the given inequality holds for the given values of \(\mathrm{x}\). $$ \frac{1}{4} x+1 / x>1 \text { for } x>2 $$

Step-by-Step Solution

Answer

The inequality \( \frac{1}{4}x + \frac{1}{x} > 1 \) holds for all \( x > 2 \) due to the nature of the upward-opening quadratic.

Step 1: Understand the Inequality

We need to show that the inequality \( \frac{1}{4} x + \frac{1}{x} > 1 \) holds true for all \( x > 2 \). This means, no matter what value greater than 2 we choose for \( x \), the left side should always be greater than 1.

Step 2: Simplify the Inequality

First, let's manipulate the inequality algebraically. Multiply through by \( x \) (keeping in mind that \( x > 2 \) implies \( x eq 0 \)) to eliminate the fraction:\[ x \left(\frac{1}{4} x + \frac{1}{x} \right) > x(1) \] \[ \frac{1}{4}x^2 + 1 > x \] This simplifies the problem to checking if this new inequality holds for \( x > 2 \).

Step 3: Rearrange the Simplified Inequality

Rearrange the inequality to bring terms to one side:\[ \frac{1}{4}x^2 - x + 1 > 0 \]We now need to show that this expression is always positive for \( x > 2 \).

Step 4: Analyze the Quadratic

Consider the quadratic expression \( \frac{1}{4}x^2 - x + 1 \). Since it is a quadratic in the form \( ax^2 + bx + c \) with \( a = \frac{1}{4} > 0 \), it opens upwards, indicating that it will form a parabola with a minimum point. We should find the roots or check using the quadratic formula or derivative methods to see where this expression exceeds zero.

Step 5: Check for Minimum or Roots

Let's check this by completing the square or finding the vertex to determine its minimum value. It's more straightforward to calculate derivatives or assess the value directly for given \( x \): For \( x > 2 \), substitute values to verify:For \( x = 2 \), \( \frac{1}{4}(2)^2 - 2 + 1 = \frac{1}{4} \times 4 - 2 + 1 = 1 - 2 + 1 = 0 \).For any value slightly more than 2 (e.g., 2.1), this will become positive as the parabola opens upwards.

Step 6: Conclusion from Analysis

Since the parabola opens upwards, and the lowest point occurs at \( x = 2 \) where the inequality holds strictly, for all \( x > 2 \), \( \frac{1}{4}x^2 - x + 1 > 0 \), ensuring \( \frac{1}{4} x + \frac{1}{x} > 1 \) is true for all \( x > 2 \).

Key Concepts

Quadratic Inequality

Quadratic inequalities involve expressions where the variable appears as a square. These inequalities take the form of a quadratic equation but use inequality signs like ">", "<", "≤", or "≥". In the given exercise, our task was to prove that \( \frac{1}{4}x + \frac{1}{x} > 1 \) holds true for all \( x > 2 \).

The first step usually involves simplifying such inequalities. We made the inequality easier to handle by multiplying through by \( x \), resulting in a simpler form \( \frac{1}{4}x^2 + 1 > x \). This reformation built a connection to quadratic expressions.

Understanding how to manage these inequalities is crucial in calculus because they often find their applications in problems needing range determinations, like the domain of functions.

Solving Inequalities

Solving inequalities means finding all the possible values of the variable that make the inequality true. Here, we dealt with \( \frac{1}{4}x^2 - x + 1 > 0 \) for \( x > 2 \).

The typical method includes rearranging the inequality to set one side to zero. This helps in identifying the condition of the expression being positive or negative.

Once rearranged, finding the roots of the expression using tools such as the quadratic formula or factorization, allows us to demarcate intervals where the inequality holds.

Interestingly, for quadratics like in this example, identifying whether the parabola opens upwards or downwards reveals the region of feasible solutions (where it exceeds zero). Solving inequalities is fundamental as it paves the way for assessing constraints and limits within calculus.

Parabolic Functions

Parabolic functions are represented by quadratic equations, forming a distinctive U-shaped curve called a parabola. The structure of a quadratic equation \( ax^2 + bx + c \) dictates the parabola's characteristics.

In our exercise, \( \frac{1}{4}x^2 - x + 1 \) describes a parabola opening upwards, as the leading coefficient \( a = \frac{1}{4} > 0 \). This property implies the chance for a minimum point on the graph.

The parabola's vertex, or turning point, is critical, as it marks the lowest or highest point of the curve. For our function, determining that the parabola opens upwards promised that beyond this minimum point (specifically for \( x > 2 \)), the function's values are positive.

Understanding how parabolas behave aids in evaluating solutions of quadratic inequalities, as it becomes clearer when and why certain interval solutions apply, directly impacting calculus problem-solving.