/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Let \(f(x)\) be a quadratic expr... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(f(x)\) be a quadratic expression possible for all real \(x\). If \(g(x)=f(x)-f^{\prime}(x)+f^{\prime \prime}(x)\), then for any real \(x\) (a) \(g(x)>0\) (b) \(g(x) \leq 0\) (c) \(g(x) \geq 0\) (d) \(g(x)<0\)

Short Answer

Expert verified
The correct answer is (c) \(g(x) \geq 0\) for any real \(x\).

Step by step solution

01

Identify the Form of f(x)

Since \(f(x)\) is a quadratic expression, it takes the general form \(f(x) = ax^2 + bx + c\), where \(a, b,\) and \(c\) are constants.
02

Calculate the First Derivative of f(x)

Find the first derivative \(f^{\prime}(x)\). Using the power rule, \(f^{\prime}(x) = \frac{d}{dx}(ax^2 + bx + c) = 2ax + b\).
03

Calculate the Second Derivative of f(x)

Find the second derivative \(f^{\prime\prime}(x)\). Derive \(f^{\prime}(x) = 2ax + b\) again to get \(f^{\prime\prime}(x) = \frac{d}{dx}(2ax + b) = 2a\).
04

Substitute Expressions into g(x)

Substitute \(f(x)\), \(f^{\prime}(x)\), and \(f^{\prime\prime}(x)\) into \(g(x)\): \[g(x) = (ax^2 + bx + c) - (2ax + b) + 2a\].
05

Simplify the Expression for g(x)

Combine like terms in the expression for \(g(x)\): \[g(x) = ax^2 + bx + c - 2ax - b + 2a = ax^2 + (b - 2a)x + (c - b + 2a)\].
06

Analyze g(x) for All Real x

Since \(ax^2 + (b - 2a)x + (c - b + 2a)\) is quadratic and \(ax^2\) always has a non-negative minimum (0 for real coefficients) when \(a > 0\), and the linear and constant terms can adjust \(b - 2a\) and \(c - b + 2a\) but do not change this minimum property, the expression \(g(x)\) can achieve values greater than or equal to zero, depending on the specific values of \(a, b,\) and \(c\). Thus, \(g(x) \geq 0\).

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Key Concepts

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

Differentiation
Differentiation is a fundamental mathematical method that helps us understand how functions change. It's like looking at a road map to see where the turns and slopes are. This process is essential in calculus and is used to find the rate at which a function is changing at any given point. When you differentiate an equation, you are essentially finding its derivative.
  • Derivatives provide insights into the behavior of functions.
  • They help identify trends, such as increasing or decreasing functions.
  • Differentiation applies rules like the power rule, product rule, and chain rule to dissect functions.
For a quadratic function, the differentiation reveals the nature of its curve, giving valuable information about how the function grows or shrinks.
First Derivative
The first derivative of a function is a primary tool for understanding its immediate behavior. For a quadratic function, expressed as \(f(x) = ax^2 + bx + c\), the first derivative, \(f^{\prime}(x)\), provides insight into the slope of the function at any given point. Using the power rule, we find:
\[f^{\prime}(x) = \frac{d}{dx}(ax^2 + bx + c) = 2ax + b\]This derivative tells us:
  • The slope of the tangent line to the function at any point \(x\).
  • Where the function is increasing or decreasing. When \(f^{\prime}(x) > 0\), the function is increasing; when \(f^{\prime}(x) < 0\), it's decreasing.
  • The points where the slope is zero, namely where the function changes from increasing to decreasing (and vice versa), indicating potential maximum or minimum points.
Thus, the first derivative is a crucial component for sketching the basic curve of a quadratic function.
Second Derivative
The second derivative gives deeper insights into the curvature or concavity of the function. It evaluates the rate of change of the first derivative. For our quadratic function, \(f^{\prime\prime}(x)\) is:
\[f^{\prime\prime}(x) = \frac{d}{dx}(2ax + b) = 2a\]What does this tell us?
  • If \(f^{\prime\prime}(x) > 0\), the function is concave up, resembling a 'U' shape. This suggests a minimum point.
  • If \(f^{\prime\prime}(x) < 0\), the function is concave down, like an upside-down 'U', indicating a maximum point.
  • For quadratic functions, if \(2a > 0\), it reveals that the entire function is shaped as a parabola opening upwards.
Thus, the second derivative is vital for understanding where the function achieves its extremes.
Quadratic Expression Analysis
Analyzing a quadratic expression involves understanding its graphical representation, roots, maximum or minimum values, and overall trajectory. Rewriting the function in standardized forms helps in revealing its characteristics.
Here, the expression \(ax^2 + (b - 2a)x + (c - b + 2a)\) demonstrates a transformed quadratic equation:
  • The component \(ax^2\) tells that the core shape is a parabola.
  • The linear term \((b - 2a)x\) shifts this parabola laterally and can influence the direction of the curve.
  • The constant \((c - b + 2a)\) indicates vertical adjustments off the \(x\)-axis.
Quadratic analysis reveals behaviors such as where the graph crosses the axes and what extreme values it might reach. This analysis helps predict outcomes for different ranges of input, making it a powerful tool in practical applications like physics and engineering.

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