Problem 15 - 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) \(x=3\) or \(x=-3\); (d) No.

Step 1: Find y when x is zero

To find the value of \(y\) when \(x = 0\) in the function \(y = f(x) = x^2 + 2\), substitute \(x = 0\) into the equation:\[ y = 0^2 + 2 = 2 \]The value of \(y\) is 2 when \(x\) is 0.

Step 2: Calculate f(3)

To calculate \(f(3)\), substitute \(x = 3\) into the function \(y = x^2 + 2\):\[ f(3) = 3^2 + 2 = 9 + 2 = 11 \]Hence, \(f(3)\) is 11.

Step 3: Find x for y value of 11

To find the values of \(x\) for which \(y = 11\), set the function equal to 11:\[ x^2 + 2 = 11 \]Subtract 2 from both sides to solve for \(x^2\):\[ x^2 = 9 \]Taking the square root of both sides, we get:\[ x = 3 \text{ or } x = -3 \]Therefore, \(x = 3\) or \(x = -3\) make \(y = 11\).

Step 4: Check possibilities for y value of 1

To see if there are values of \(x\) that make \(y = 1\), set the function equal to 1:\[ x^2 + 2 = 1 \]Subtract 2 from both sides to solve for \(x^2\):\[ x^2 = -1 \]As \(x^2\) cannot be negative for real numbers, there is no real value of \(x\) that gives \(y = 1\).

Key Concepts

Understanding Quadratic Functions

A quadratic function is a type of function where the highest degree of the variable is two. It is generally written in the form \( f(x) = ax^2 + bx + c \), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). In the original exercise, we explored the quadratic function \( f(x) = x^2 + 2 \).

Quadratic functions graphically represent parabolas. A parabola can open upwards or downwards, depending on the sign of the constant \(a\). Here, since \(a = 1\) (which is positive), the parabola opens upwards. The constant \(c\) controls where the graph crosses the y-axis. In our case, \(c = 2\), meaning the parabola crosses the y-axis at (0,2).

Studying quadratic functions is crucial as they're foundational in algebra. These functions model various real-world problems, such as projectile motion in physics.

Evaluating a Function

Function evaluation is the process of calculating the output of a function for a certain input value. To evaluate a function like \( f(x) = x^2 + 2 \), we replace \(x\) with a specific number and apply the given operations.

For example, to find \( f(3) \), substitute 3 for \(x\) in the equation. This becomes \( 3^2 + 2 = 9 + 2 = 11 \). Thus, \( f(3) = 11 \). This step-by-step calculation process allows us to assess the function's value at any given point. Function evaluation is a fundamental concept useful across mathematics, enabling us to analyze and predict function behavior for specific inputs.

It is also essential when solving equations, as it helps in determining whether certain values of the variable result in specific outputs.

Solving Equations Involving Quadratic Functions

Solving equations entails finding values of the variable that make an equation true. In the context of quadratic functions, this can involve setting the quadratic expression equal to a number and solving for the variable.

For instance, if \( f(x) = x^2 + 2 \) and we want \( f(x) = 11 \), solve \( x^2 + 2 = 11 \). Subtract 2 from both sides to simplify: \( x^2 = 9 \), then take the square root of both sides. We find \( x = 3 \) or \( x = -3 \). This means both values satisfy the original equation.

However, not all quadratic equations yield real solutions. If the set equation produces a negative number under the square root, like \( x^2 = -1 \), there are no real solutions. This highlights the importance of understanding the nature of quadratic solutions, which can be either real or imaginary depending on the expression. Mastery of this skill helps in solving a wide range of mathematical and real-world problems.