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Let \(f(x)=a x^{2}+b x+c,\) where \(a<0 .\) Show that \(\frac{f\left(x_{1}\right)+f\left(x_{2}\right)}{2} \leq f\left(\frac{x_{1}+x_{2}}{2}\right)\)

Short Answer

Expert verified
The inequality holds by concavity, as the function is concave down for \( a < 0 \).

Step by step solution

01

Understand the Problem

We need to prove that for the quadratic function \( f(x) = ax^2 + bx + c \) where \( a < 0 \), the average value of the function at two points \( x_1 \) and \( x_2 \) is less than or equal to the function's value at the midpoint of these two points.
02

Consider the Function Type

Since \( a < 0 \), the quadratic function \( f(x) = ax^2 + bx + c \) is a concave down parabola. The statement we need to prove relates to the concept of concavity.
03

Apply Jensen's Inequality

Jensen's inequality applies to concave functions. It states that \( f(\lambda x_1 + (1 - \lambda)x_2) \geq \lambda f(x_1) + (1 - \lambda) f(x_2) \) for \( 0 \leq \lambda \leq 1 \). For our case, use \( \lambda = 0.5 \).
04

Calculate the Midpoint Value

Calculate the midpoint: \( m = \frac{x_1 + x_2}{2} \). Evaluate the function at this midpoint: \( f(m) = a \left( \frac{x_1 + x_2}{2} \right)^2 + b \left( \frac{x_1 + x_2}{2} \right) + c \).
05

Use Jensen's Inequality in Our Case

By Jensen's inequality for \( \lambda = 0.5 \):\[ f\left(\frac{x_1 + x_2}{2}\right) \geq \frac{1}{2}f(x_1) + \frac{1}{2}f(x_2) \]This inequality holds since \( f \) is concave down due to \( a < 0 \).
06

Conclude the Inequality

The expression from the previous step directly shows \( \frac{f(x_1) + f(x_2)}{2} \leq f\left( \frac{x_1 + x_2}{2} \right) \). This completes the proof.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Concavity
Concavity is a concept that shows whether a curve bends upwards or downwards. In mathematical terms, when a function is "concave down," it means that as you look at it from left to right, it opens downward like an upside-down bowl. This is represented in a quadratic function by the sign of the coefficient "\( a \)" in the equation \( f(x) = ax^2 + bx + c \). If \( a < 0 \), the parabola is concave down.

Understanding concavity is crucial in analyzing functions, especially quadratic ones. For instance, when we say \( f(x) = ax^2 + bx + c \) is concave down, it ensures that any line drawn between any two points on this curve will sit above or touch the curve. This property plays a critical role when evaluating inequalities in quadratic functions and is at the heart of concepts like Jensen's Inequality.
Jensen's Inequality
Jensen's Inequality is a powerful mathematical tool used to relate the values of a concave function at different points. It is especially useful when dealing with averages. For a concave function \( f \), Jensen’s Inequality states that:
  • \( f(\lambda x_1 + (1 - \lambda)x_2) \geq \lambda f(x_1) + (1 - \lambda) f(x_2) \)
where \( 0 \leq \lambda \leq 1 \). This means if you take a linear combination of two points and evaluate the function at this combination, the result will be higher than or equal to the weighted average of the function's value at those two points.

In the context of the quadratic function \( f(x) = ax^2 + bx + c \) where \( a < 0 \), Jensen’s Inequality helps us prove that the value at the midpoint will indeed be greater than or equal to the average of the values at the endpoints \( x_1 \) and \( x_2 \). By choosing \( \lambda = 0.5 \), the inequality simplifies, demonstrating the function's average value characteristics over these points on a concave down parabola.
Parabolas
Parabolas are fundamental shapes in mathematics that result from plotting quadratic functions. They appear in the form \( f(x) = ax^2 + bx + c \) and have characteristic curves known as U-shapes or inverted U-shapes. The direction they open, either up or down, depends on the coefficient \( a \).

When \( a > 0 \), the parabola opens upwards, and when \( a < 0 \), as in our exercise, it opens downwards. This dictates the function's concavity and consequently influences how it behaves over an interval. With \( a < 0 \), the parabola represents a maximum point at its vertex, and every other point on the curve lies below this peak. Such characteristics are essential in solving inequality problems as they display how the function's value changes across the domain, particularly showcasing where it peaks and descends ensuring that our inequality criteria using Jensen's Inequality are satisfied.

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Most popular questions from this chapter

Use a graphing utility to graph each function and then apply the horizontal line test to see whether the function is one-to-one. $$y=2 x^{3}+x^{2}$$

For Exercises 58 and \(59,\) assume that \((a, b)\) is a point on the graph of \(y=f(x),\) and specify the corresponding point on the graph of each equation. [For example, the point that corre- sponds to \((a, b)\) on the graph of \(y=f(x-1)\) is (\(a+1, b).] (a) \)y=f(x-3)\( (b) \)y=f(x)-3\( (c) \)y=f(x-3)-3\( (d) \)y=-f(x)$

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Use the horizontal line test to determine whether the function is one-to-one (and therefore has an inverse ). (You should be able to sketch the graph of each function on your own, without using a graphing utility.) $$g(x)=\left\\{\begin{array}{ll}x^{2} & \text { if }-1 \leq x<0 \\\x^{2}+1 & \text { if } x \geq 0\end{array}\right.$$

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