Question

Consider the following function. $$f(x)=\left\\{\begin{array}{rlr} 3 x-2 & \text { if } & x<2 \\ x^{2}+5 & \text { if } & 2 \leq x<10 \\ 3 & \text { if } & x \geq 10 \end{array}\right. $$ Which expression(s) should be used to find the \(y\) -intercept? (a) \(3 x-2\) (b) \(x^{2}+5\) (c) 3 (d) all three

Short answer

(a) 3x - 2

Step-by-step solution

Understand the Function and Intervals

The function is defined in three different intervals: 1. For \(x < 2\), the expression is \(3x - 2\). 2. For \(2 \leq x < 10\), the expression is \(x^2 + 5\). 3. For \(x \geq 10\), the expression is \(3\). To find the \(y\)-intercept, determine what happens when \(x = 0\).

Identify the Applicable Interval for \(x = 0\)

Since \(0 < 2\), use the expression defined for the interval \(x < 2\).

Apply the Expression for \(x < 2\) at \(x = 0\)

The expression for \(x < 2\) is \(3x - 2\). Substitute \(x = 0\) into this expression: \[ f(0) = 3(0) - 2 = -2 \]

Conclusion

The function value at \(x = 0\) is \(-2\). Therefore, the expression used to find the \(y\)-intercept is \(3x - 2\).

Key concepts

y-intercept

To find the y-intercept of a function, we need to evaluate the function at x = 0. This intersection point is where the graph of the function crosses the y-axis. In the given piecewise function, there are different expressions for different intervals of x. To pinpoint which expression to use, we check the interval that includes x = 0. Since 0 is less than 2, we use the expression for that interval: 3x - 2. Evaluating this at x = 0, we get: \[ f(0) = 3(0) - 2 = -2 \]Hence, the y-intercept is -2.

intervals

Intervals specify the segments within which each piece of the piecewise function is applicable. In this function:

• For x < 2, the function is 3x - 2.
• For 2 ≤ x < 10, the function is x^2 + 5.
• For x ≥ 10, the function is 3.

Each interval confines the behavior of the function to a predetermined form, ensuring different rules for different ranges of x.

function evaluation

Function evaluation means finding the function's output for specific values of x. For piecewise functions, this process involves:

• Identifying which interval the x-value falls into.
• Using the corresponding expression.
• Substituting the specific x-value into the chosen formula.

For example, evaluating the function at x = 5, which falls in the interval 2 ≤ x < 10, we use x^2 + 5: \[ f(5) = 5^2 + 5 = 25 + 5 = 30 \]

continuity

Continuity in a function means there are no jumps, breaks, or holes in its graph. For piecewise functions, we check continuity at the boundaries between intervals. In this function, those critical points are x = 2 and x = 10. To ensure continuity, the function's value should be the same when approaching these points from both sides.

• At x = 2, evaluate both 3x - 2 and x^2 + 5. \[ 3(2) - 2 = 4 \]and\[ 2^2 + 5 = 9 \]. Since 4 ≠ 9, the function is not continuous at x = 2.
• At x = 10, evaluate both x^2 + 5 and 3. \[ 10^2 + 5 = 105 \]and\[ 3 \]. Since 105 ≠ 3, the function is not continuous at x = 10.

So, the function has discontinuities at x = 2 and x = 10.