Question

the graph of a function (y = f(x)) is shown. make a sketch of the antiderivative (f), given that (f) is continuous and (f(0) = -2).

Explanation:

Analyze the given function \(f(x)\)

Using the Antiderivatives knowledge point
The graph of \(y = f(x)\) consists of constant horizontal segments and isolated points:

• For \(0 \le x < 2\): \(f(x) = 1\)
• At \(x = 2\): \(f(2) = -2\) (isolated point)
• For \(2 < x < 5\): \(f(x) = -2\)
• At \(x = 5\): \(f(5) = 0\) (isolated point)
• For \(5 < x \le 8\): \(f(x) = 0\)
• Isolated points at \((3.6, 4.5)\) and \((7.7, -4.3)\) do not affect the integral of the piecewise continuous function.

Determine the values of \(F(x)\) using integration

Using the Antiderivatives knowledge point
We are given that \(F(0) = -2\) and \(F\) is continuous. Since \(F'(x) = f(x)\) almost everywhere:

• For \(0 \le x \le 2\):

$$ F(x) = F(0) + \int_{0}^{x} 1 \, dt = -2 + x $$

At \(x = 2\), \(F(2) = -2 + 2 = 0\).

• For \(2 \le x \le 5\):

$$ F(x) = F(2) + \int_{2}^{x} -2 \, dt = 0 - 2(x - 2) = -2x + 4 $$

At \(x = 5\), \(F(5) = -2(5) + 4 = -6\).

• For \(5 \le x \le 8\):

$$ F(x) = F(5) + \int_{5}^{x} 0 \, dt = -6 $$

At \(x = 8\), \(F(8) = -6\).

Identify key coordinates for sketching \(F(x)\)

Using the Antiderivatives knowledge point
The continuous antiderivative \(F(x)\) is composed of connected linear segments:

• Segment 1: From \((0, -2)\) to \((2, 0)\) with slope \(1\).
• Segment 2: From \((2, 0)\) to \((5, -6)\) with slope \(-2\).
• Segment 3: From \((5, -6)\) to \((8, -6)\) with slope \(0\).

Answer:

To sketch the continuous antiderivative \(F(x)\) given \(F(0) = -2\), plot the following connected line segments:

1. A straight line segment from \((0, -2)\) to \((2, 0)\) (slope is \(1\)).
2. A straight line segment from \((2, 0)\) to \((5, -6)\) (slope is \(-2\)).
3. A horizontal line segment from \((5, -6)\) to \((8, -6)\) (slope is \(0\)).