Problem 21 - The Calculus with Analytic Geometry

Question

\(f(x)= \begin{cases}3 x+5 & \text { if } x<-1 \\ x^{2}+1 & \text { if }-1 \leq x<2 \\ 7-x & \text { if } 2 \leq x\end{cases}\)

Step-by-Step Solution

Answer

The function f(x) is defined piecewise with different expressions for different intervals of x.

Step 1 - Understand the Piecewise Function

Identify the different cases for the function: 1. When \text{\(x < -1\)}, the function is defined as \text{\(f(x) = 3x + 5\)}.2. When \text{\(-1 \text{ ≤ } x < 2\)}, the function is defined as \text{\(f(x) = x^2 + 1\)}.3. When \text{\(x \text{ ≥ } 2\)}, the function is defined as \text{\(f(x) = 7 - x\)}.

Step 2 - Evaluate the Function at Different Points

For a better understanding, evaluate the function at key points: - For \text{\(x = -2\)} (since \text{\(-2 < -1\)}): \text{\(f(-2) = 3(-2) + 5 = -6 + 5 = -1\)}- For \text{\(x = -1\)} (since \text{\(-1 \text{ ≤ } -1 < 2\)}): \text{\(f(-1) = (-1)^2 + 1 = 1 + 1 = 2\)}- For \text{\(x = 1\)} (since \text{\(-1 \text{ ≤ } 1 < 2\)}): \text{\(f(1) = 1^2 + 1 = 1 + 1 = 2\)}- For \text{\(x = 2\)} (since \text{\(2 \text{ ≥ } 2\)}): \text{\(f(2) = 7 - 2 = 5\)}- For \text{\(x = 3\)} (since \text{\(3 \text{ ≥ } 2\)}): \text{\(f(3) = 7 - 3 = 4\)}.

Step 3 - Sketch the Graph

Draw the graph by plotting the points and connecting them according to the function definitions: 1. For \text{\(x < -1\)} (\text{\(f(x) = 3x + 5\)}), draw a line through points like \text{(-2, -1)} and approaching x = -1.2. For \text{\(-1 \text{ ≤ } x < 2\)} (\text{\(f(x) = x^2 + 1\)}), draw a parabola opening upwards, starting at \text{(-1, 2)} and ending at (just before) \text{(2, 5)}.3. For \text{\(x \text{ ≥ } 2\)} (\text{\(f(x) = 7 - x\)}), draw a line starting from point (2, 5) and continuing through, such as \text{(3, 4)}.

Key Concepts

Graphing Piecewise Functions

Graphing a piecewise function can initially seem challenging, but breaking it down into steps makes it manageable. Start by identifying each segment of the function. For example, the piecewise function provided is:
\(f(x)= \begin{cases}3 x+5 & \text { if } x<-1 \ x^{2}+1 & \text { if }-1 \leq x<2 \ 7-x & \text { if } 2 \leq x\end{cases}\).

• For \(x < -1\), plot the linear function \(f(x) = 3x + 5\). This line has points such as (-2, -1) and approaching x = -1.
• Next, for \(-1 \leq x < 2\), plot the parabola \(f(x) = x^2 + 1\), starting from (-1, 2) and ending just before (2, 5).
• Finally, for \(x \geq 2\), plot the line \(f(x) = 7 - x\) starting at (2, 5) and continuing through (3, 4).

When drawing these graphs, be sure to use open or closed circles to indicate whether the endpoints are included in that segment. This will help accurately represent the function at those boundaries.

Evaluating Piecewise Functions

To evaluate piecewise functions, determine which segment applies to the given value of \(x\). For instance, consider our piecewise function:
– For \(x = -2\), since \(-2 < -1\), use \(f(x) = 3x + 5\):
\(f(-2) = 3(-2) + 5 = -6 + 5 = -1\).
– For \(x = -1\), since \(-1 \leq -1 < 2\), use \(f(x) = x^2 + 1\):
\(f(-1) = (-1)^2 + 1 = 1 + 1 = 2\).
– For \(x = 1\), since \(-1 \leq 1 < 2\), use \(f(x) = x^2 + 1\):
\(f(1) = 1^2 + 1 = 1 + 1 = 2\).
– For \(x = 2\), since \(2 \geq 2\), use \(f(x) = 7 - x\):
\(f(2) = 7 - 2 = 5\).
– For \(x = 3\), since \(3 \geq 2\), use \(f(x) = 7 - x\):
\(f(3) = 7 - 3 = 4\).
Thus, each evaluation is straight-forward once you identify the correct equation for the specific interval of the piecewise function.

Piecewise Function Continuity

Continuity in piecewise functions is about ensuring there are no breaks or jumps in the graph. To check for continuity, look at the function at the boundaries of the intervals. For the given function:
\(f(x)= \begin{cases}3 x+5 & \text { if } x<-1 \ x^{2}+1 & \text { if }-1 \leq x<2 \ 7-x & \text { if } 2 \leq x\end{cases}\),
you need to check if the function's values match at these transition points.

• At \(x = -1\), check transition between \(f(x) = 3x + 5\) and \(f(x) = x^2 + 1\).
\[ \text{Limit from the left as } x \to -1^- = 3(-1) + 5 = 2, \text{ Right value as } x = -1: (-1)^2 + 1 = 2 \rightarrow \text{No jump at } x = -1 \]
• At \(x = 2\), check transition between \(f(x) = x^2 + 1\) and \(f(x) = 7 - x\).
\[ \text{Limit from the left as } x \to 2^-: 2^2 + 1 = 5, \text{ Right value as } x = 2: 7 - 2 = 5 \rightarrow \text{No jump at } x = 2 \]

Because the function values match, we confirm that there are no jumps, making the function continuous at these points. Ensuring continuity involves checking these key transitions to see if the left-hand and right-hand limits, along with the function's value at those points, match.