Problem 44 - Calculus Single Variable

Question

Find and test the critical points of the functions in Exercises 43-47. $$ f(x)=|x-7|+x^{2} $$

Step-by-Step Solution

Answer

Critical points occur at \( x = 0.5 \) and need to verify the behavior around \( x = 7 \).

Step 1: Identify the function's components

The function is given as \( f(x) = |x-7| + x^2 \). This function is a combination of an absolute value function and a quadratic function. The absolute value component, \(|x-7|\), can create points where the function changes from increasing to decreasing or vice versa at \(x = 7\).

Step 2: Split the function into cases

Because of the absolute value, we divide the function into two cases:1. When \( x \geq 7 \), the absolute value \(|x-7| = x - 7\), so \( f(x) = (x - 7) + x^2 = x^2 + x - 7 \).2. When \( x < 7 \), the absolute value \(|x-7| = -(x - 7) = 7 - x\), so \( f(x) = (7 - x) + x^2 = x^2 - x + 7 \).

Step 3: Find derivative for each case

Calculate the derivative of each case.For \( x \geq 7 \):\( f'(x) = 2x + 1 \).For \( x < 7 \):\( f'(x) = 2x - 1 \).

Step 4: Find critical points by setting the derivatives to zero

Critical points occur where the derivative is zero or undefined.For \( x \geq 7 \):Set \( 2x + 1 = 0 \) which gives \( x = -\frac{1}{2} \). Since \( -\frac{1}{2} < 7 \), this value doesn't belong to this interval.For \( x < 7 \):Set \( 2x - 1 = 0 \) which gives \( x = \frac{1}{2} \). Since \( \frac{1}{2} < 7 \), this point is valid for this interval.

Step 5: Check the critical points and at the point of transition

Check the function at the transition point \( x = 7 \) where the form of the function changes.Evaluate \( f(x) \) and its derivative on either side of \( x = 7 \):- For \( x = 7 \), \( f'(x) \) changes from \( 2x - 1 \) to \( 2x + 1 \). Neither is zero at \( x = 7 \), and both are defined.- Check \( f(\frac{1}{2}) \) and \( f(7) \) if they are critical points by substituting back in the original equation.

Step 6: Calculate function values for critical points

For \( x = \frac{1}{2} \), calculate \( f\left( \frac{1}{2} \right) = \left|\frac{1}{2} - 7\right| + \left( \frac{1}{2} \right)^2 = \frac{13}{2} + \frac{1}{4} = \frac{27}{4} \).For \( x = 7 \), calculate \( f(7) = |7-7| + 7^2 = 0 + 49 = 49 \).

Key Concepts

Absolute Value Functions

An absolute value function involves the absolute value expression, like \(|x-7|\) in our exercise. This function represents the distance of a number from zero on the number line, making it always non-negative.
When dealing with absolute value functions, the key feature is their V-shape.
It occurs because the function changes direction at the point where the expression inside the absolute value equals zero. For example, in \(|x-7|\), the critical turning point is at \(x = 7\).

• When \(x\) is greater than or equal to the value inside the absolute value (\(x-7\) set to zero), the expression \(|x-7|\) simplifies to \(x-7\).
• When \(x\) is less than this point (\(7\)), the expression becomes \(7-x\), reversing the sign.

It’s essential to treat absolute value functions in a piecewise manner due to their different behaviors in different intervals.

Quadratic Functions

Quadratic functions are polynomial functions of the form \(ax^2 + bx + c\) where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). The function part \(x^2\) of our main function is a standard quadratic function.
These functions graph to a parabola, opening upwards when \(a > 0\) and downwards when \(a < 0\).

• The vertex of the parabola represents the highest or lowest point on the graph.
• The axis of symmetry can be calculated using \(x = -\frac{b}{2a}\).

In the context of critical points, quadratics typically have either a maximum or minimum point, which can be assessed through their derivative analysis. In combination with absolute values, it contributes to identifying intervals of increase and decrease.

Derivative Calculation

The derivative of a function measures the rate at which the function's value changes as its input changes.
Calculating the derivative is crucial to finding critical points because they are where the derivative is zero or undefined.
For our function, different rules apply based on the pieces defined by the absolute value component:

• When \(x \geq 7\), the function becomes \(f(x) = x^2 + x - 7\). Its derivative, \(f'(x) = 2x + 1\), is found using simple power and sum rules of differentiation.
• When \(x < 7\), the function turns into \(f(x) = x^2 - x + 7\), resulting in the derivative \(f'(x) = 2x - 1\).

Each piece's derivative will help determine where the slope is zero, indicating potential critical points.

Piecewise Functions

A piecewise function is defined by different expressions or rules over different intervals of the input variable. This approach allows complex functions to be analyzed in simpler parts, as seen with \(f(x) = |x-7| + x^2\).
In this example, splitting the function based on the absolute value gives:

• For \(x \geq 7\): \(f(x) = x^2 + x - 7\).
• For \(x < 7\): \(f(x) = x^2 - x + 7\).

Each piece can be treated separately to calculate derivatives and critical points. Piecewise definitions are very helpful when you have changes in behavior at specific values, providing clarity in situations involving absolute value functions or within different intervals of polynomials.