Problem 96 - Calculus Volume 1

Question

For each of the piecewise defined functions, a. evaluate at the given values of the independent variable and b. sketch the graph. \(h(x)=\left\\{\begin{array}{c}x+1, x \leq 5 \\ 4, x>5\end{array} ; h(0) ; h(\pi) ; h(5)\right.\)

Step-by-Step Solution

Answer

h(0)=1, h(\pi)\approx4.14, h(5)=6. Graph transitions from a line to a horizontal at x=5.

Step 1: Evaluate h(0)

To find \(h(0)\), we determine which part of the piecewise function to use. Since \(0 \leq 5\), we use \(h(x) = x + 1\). Plug \(0\) into this function: \(h(0) = 0 + 1 = 1\).

Step 2: Evaluate h(\pi)

Determine which condition \(\pi\) falls under. Since \(\pi \approx 3.14\) and \(3.14 \leq 5\), we use \(h(x) = x + 1\). Substitute \(\pi\) to get \(h(\pi) = \pi + 1\). The value of \(h(\pi)\) is approximately \(4.14\).

Step 3: Evaluate h(5)

Check which part of the function to use for \(h(5)\). Since \(5 \leq 5\), we use \(h(x) = x + 1\). Replace \(x\) with \(5\): \(h(5) = 5 + 1 = 6\).

Step 4: Sketch the Graph

Start by sketching \(y = x + 1\) for \(x \leq 5\). This is a straight line with a slope of 1 that passes through \((0,1)\) and ends at \((5,6)\). For \(x > 5\), draw a horizontal line at \(y = 4\) starting from \((5,4)\) rightwards. Mark an open circle at \((5,4)\) to indicate this point is excluded from the line \(y=4\) but included in \((5,6)\). The sketch should show a continuous transition from \(x=5\) between these pieces.

Key Concepts

Function Evaluation

When working with piecewise functions, evaluating the function at specific points is key. A piecewise function is defined by different expressions over different intervals of the independent variable, often noted as \(x\). In our exercise, we have:

• \(h(x) = x + 1\) for \(x \leq 5\)
• \(h(x) = 4\) for \(x > 5\)

To evaluate \(h(x)\) at certain values, such as \(h(0), h(\pi), \text{and } h(5)\), identify which rule of the piecewise function applies.
For \(h(0)\), since \(0 \leq 5\), we use \(h(x) = x + 1\). Therefore, \(h(0) = 0 + 1 = 1\).
Similarly, as \(\pi \approx 3.14\) and is also \(\leq 5\), \(h(\pi) = \pi + 1\), which approximates to \(4.14\).
Finally, for \(h(5)\), since \(5 \leq 5\), using \(h(x) = x + 1\) gives \(h(5) = 6\).
Understanding which part of the piecewise function to use is crucial in function evaluation. It requires careful checking of the conditions defining each piece.

Graph Sketching

Graphing a piecewise function involves plotting each piece according to its specific rule and interval. Start by considering each segment individually:

• The first piece \(y = x + 1\) applies for \(x \leq 5\).
• The second piece \(y = 4\) is valid for \(x > 5\).

To sketch, begin with the segment \(y = x + 1\). It forms a straight line with a slope of 1. Start at \((0,1)\), pass through \((5,6)\), and draw a solid line indicating the inequality \(x \leq 5\).
Next, for the segment \(y = 4\), draw a horizontal line starting just right of \(x = 5\). Use an open circle at \((5,4)\) to show this point isn't included on this line, but the endpoint \((5,6)\) from the previous segment is included.
These sketches ensure that the piecewise function is accurately represented. The transition at \(x = 5\) must be clear, with one piece ending where the other begins.

Calculus Problems

Piecewise functions often pose interesting challenges in calculus, especially when considering continuity and differentiability. A function's continuity at a point means its value from the left matches its value from the right.
In our exercise, examine the continuity at \(x = 5\). For this to hold, both \(h(x) = x + 1\) for \(x \leq 5\) and \(h(x) = 4\) for \(x > 5\) must meet at the same \(y\)-value at \(x = 5\). Here, \(h(5)=6\), but the next piece starts at \(y=4\), so it's not continuous at this point.

• The function jumps from \(y=6\) to \(y=4\).
• To address differentiability, note that a corner exists at \(x = 5\), hence \(h(x)\) is non-differentiable there.

Calculus with piecewise functions requires extra care in analyzing these points of transition, observing how the pieces connect, and understanding their implications on the function's behavior at those points.