Problem 10 - The Calculus with Analytic Geometry

Question

\(f(x)= \begin{cases}x^{2} & \text { if } x<1 \\ x^{3}-4 x^{2}+7 x-3 & \text { if } x \geq 1\end{cases}\)

Step-by-Step Solution

Answer

Use \[f(x) = x^{2}] for x < 1 and \[f(x) = x^{3} - 4x^{2} + 7x - 3] for x ≥ 1.

Step 1: Identify the piecewise function

The function is defined as a piecewise function with two parts:For values of x less than 1, use: \[f(x) = x^{2}\] For values of x greater than or equal to 1, use: \[f(x) = x^{3} - 4x^{2} + 7x - 3\]

Step 2: Evaluate the function for x < 1

To find the value of the function for \(x < 1\), use the equation: \[f(x) = x^{2}\]For example, if \(x = 0.5\), then \[f(0.5) = (0.5)^{2} = 0.25\]

Step 3: Evaluate the function for x ≥ 1

To find the value of the function for \(x \text{ ≥ } 1\), use the equation: \[f(x) = x^{3} - 4x^{2} + 7x - 3\] For example, if \(x = 2\), then \[f(2) = 2^{3} - 4(2)^{2} + 7(2) - 3 = 8 - 16 + 14 - 3 = 3\]

Key Concepts

Function Evaluation

Understanding piecewise functions is crucial for evaluating them correctly. A piecewise function is defined by different expressions based on the value of the input, x.
In this example, the function is:
1. If x < 1, use the equation: \( f(x) = x^2 \)
2. If x \( \textgreater \) or equal to 1, use the equation: \( f(x) = x^3 - 4x^2 + 7x - 3 \)
To evaluate the function at a specific point, first determine which part of the piecewise function to use. For instance:
If x is 0.5:
- Since 0.5 < 1, use \( f(x) = x^2 \)
- Substitute 0.5 for x: \( f(0.5) = (0.5)^2 = 0.25 \)
If x is 2:
- Since 2 \( \textgreater \) or equal to 1, use \( f(x) = x^3 - 4x^2 + 7x - 3 \)
- Substitute 2 for x: \( f(2) = 2^3 - 4(2)^2 + 7(2) - 3 = 8 - 16 + 14 - 3 = 3\)
Understanding which part of the function to use is key for accurate evaluations.

Continuity

Continuity in a function means that you can draw the graph without lifting your pen.
For piecewise functions, checking continuity involves ensuring that the function values match up at the boundaries of the pieces.
In this example, let's check continuity at x = 1.
1. Evaluate the left-hand limit as x approaches 1 from the left (\textless 1):
- \( \text{Since } x < 1, f(x) = x^2\)
- Substitute 1 for x: \( f(1^-) = 1^2 = 1 \)
2. Evaluate the right-hand limit as x approaches 1 from the right (\(\textgreater \)):
- Since x \( \text{≥ 1,} f(x) = x^3 - 4x^2 + 7x - 3 \)
- Substitute 1 for x: \( f(1^+) = 1^3 - 4(1)^2 + 7(1) - 3 = 1 - 4 + 7 - 3 = 1 \)
Since both limits equal 1, the function is continuous at x = 1.
Continuity means no breaks in the graph, making it easier to understand and analyze.

Algebraic Expressions

Algebraic expressions are mathematical phrases that include numbers, variables, and operations.
In piecewise functions, you'll often encounter different algebraic expressions.
For the function in our example:
1. For \( x < 1 \), the expression is a simple quadratic: \( f(x) = x^2 \)
- This is an easily recognizable form, representing a parabola opening upwards.
2. For \( x \textgreater or equal to 1 \), the expression is a cubic polynomial: \( f(x) = x^3 - 4x^2 + 7x - 3 \)
- This expression combines multiple types of terms: cubic (\textsuperscript{cubed}), quadratic (\textsuperscript{squared}), linear (x), and a constant term (-3).
To simplify and manage these expressions:
- Identify the degree of the polynomial, which tells you the highest power of x.
- Combine like terms if needed.
- Use proper substitution for function evaluation.
Understanding these expressions will help you manipulate and graph the function correctly.