Problem 12 - Applied Calculus

Question

Let \(y=f(x)=x^{2}+2\). (a) Find the value of \(y\) when \(x\) is zero. (b) What is \(f(3)\) ? (c) What values of \(x\) give \(y\) a value of 11 ? (d) Are there any values of \(x\) that give \(y\) a value of 1 ?

Step-by-Step Solution

Answer

(a) 2, (b) 11, (c) \(-3, 3\), (d) No real solutions.

Step 1: Substitute x into f(x) for part (a)

To find the value of \( y \) when \( x = 0 \), substitute \( x = 0 \) in the function. Calculate \( f(0) = 0^{2} + 2 \).

Step 2: Calculate f(0)

The calculation yields \( f(0) = 0 + 2 = 2 \).

Step 3: Substitute x into f(x) for part (b)

To find \( f(3) \), substitute \( x = 3 \) in the equation \( f(x) = x^{2} + 2 \).

Step 4: Calculate f(3)

Perform the calculation: \( f(3) = 3^{2} + 2 = 9 + 2 = 11 \).

Step 5: Solve f(x) = 11 for x for part (c)

Set up the equation \( x^{2} + 2 = 11 \). Subtract 2 from both sides to get \( x^{2} = 9 \).

Step 6: Solve x^2 = 9

Find the square root of both sides to get \( x = \pm 3 \).

Step 7: Solve f(x) = 1 for x for part (d)

Set up the equation \( x^{2} + 2 = 1 \). Subtracting 2 from both sides results in \( x^{2} = -1 \).

Step 8: Determine solution for x^2 = -1

Because the square root of a negative number is not a real number, there are no real solutions for \( x \) such that \( y = 1 \).

Key Concepts

Function Evaluation

To understand a quadratic function like \( f(x) = x^2 + 2 \), it is helpful to know how to evaluate the function at given points. Function evaluation involves plugging a specific value of \( x \) into the function and solving for \( y \). Let's break this down into manageable steps.

• Substitute the Value: Begin by replacing \( x \) in the function \( f(x) \) with the given value. For instance, if \( x = 0 \), substitute 0 in place of \( x \) in the expression.
• Compute the Result: Simplify the equation by performing the arithmetic required by the function. For \( f(x) = x^2 + 2 \), if \( x = 0 \), the arithmetic would be \( 0^2 + 2 = 2 \).
• Interpret the Output: The result from your calculation is the value of \( y \) for the specific \( x \) you chose. Hence, when \( x = 0 \), \( y = 2 \).

Evaluating functions gives you insight into how the function behaves at certain points, allowing you to sketch or analyze its graph more effectively.

Solving Equations

Solving equations is a key step in finding which values of \( x \) give us specific outputs \( y \) in a quadratic function. For example, if you want to know which \( x \) values result in \( f(x) = 11 \), you set the function equal to 11 and solve for \( x \). Here’s how you can approach it:

• Set the Equation: Replace \( f(x) \) with the desired output value. Transform \( f(x) = x^2 + 2 \) to \( x^2 + 2 = 11 \).
• Isolate the Quadratic Term: Simplify to find \( x^2 \) by subtracting any constant from both sides, yielding \( x^2 = 9 \).
• Find Solutions: Take the square root of both sides to solve for \( x \). Remember, the square root of a number has both a positive and negative solution, so \( x = \pm 3 \).

By following these steps, you can solve quadratic equations and find all possible x-values that give you a specified \( y \). This technique is crucial for fully understanding the function's behavior.

Real and Complex Numbers

When solving quadratic equations, you'll encounter both real and complex numbers. Real numbers can be found on the number line, while complex numbers include imaginary numbers. Imaginary numbers arise when taking the square root of a negative number.

• Real Numbers: If the equation you're solving, such as \( x^2 = 9 \), results in a real square root, the solutions are real numbers. For \( x^2 = 9 \), the solutions \( x = 3 \) and \( x = -3 \) are real.
• Complex Numbers: If your equation involves \( x^2 = -1 \), you cannot find a real solution because the square root of a negative is not a real number. Instead, you'd use \( i \), the imaginary unit, defined by \( i^2 = -1 \). Hence, the solutions are complex numbers \( x = i \) and \( x = -i \).

Understanding these concepts helps you work with and solve a broader range of equations, incorporating both real and complex solutions.