Question

the graph of $f$, the derivative of $f$, is shown in the graph below. if $f(0) = 4$, what is the value of $f(-3)$?

Explanation:

Step1: Recall the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that \( f(b) - f(a) = \int_{a}^{b} f'(x) \, dx \). We want to find \( f(-3) \), and we know \( f(0) = 4 \). So we can rewrite the formula as \( f(-3) - f(0) = \int_{-3}^{0} f'(x) \, dx \), which implies \( f(-3) = f(0) + \int_{-3}^{0} f'(x) \, dx \). But since \( \int_{-3}^{0} f'(x) \, dx = -\int_{0}^{-3} f'(x) \, dx \), we can also think of it as \( f(-3) = f(0) - \int_{0}^{-3} f'(x) \, dx \), or more simply, we can compute the integral from \( -3 \) to \( 0 \) of \( f'(x) \) and use \( f(0) = 4 \) to find \( f(-3) \).

Step2: Analyze the graph of \( f' \) from \( x = -3 \) to \( x = 0 \)

Looking at the graph of \( f' \):
- From \( x = -3 \) to \( x = -2 \): Wait, actually, from \( x = -3 \) to \( x = 0 \), let's break down the graph. Wait, the graph of \( f' \) from \( x = -9 \) to \( x = -6 \) is a line from \( (-9, 0) \) to \( (-6, 4) \), then from \( x = -6 \) to \( x = -3 \) it's a horizontal line at \( y = 4 \), then from \( x = -3 \) to \( x = 0 \) it's a line from \( (-3, 4) \) to \( (0, 0) \). Wait, let's confirm the shape. From \( x = -3 \) to \( x = 0 \), the graph of \( f' \) is a line segment from \( (-3, 4) \) to \( (0, 0) \). So the area under \( f' \) from \( x = -3 \) to \( x = 0 \) is the area of a triangle? Wait, no, from \( x = -3 \) to \( x = 0 \), the function \( f'(x) \) is a line with slope \( \frac{0 - 4}{0 - (-3)} = -\frac{4}{3} \), but actually, the graph from \( x = -3 \) to \( x = 0 \) is a line from \( (-3, 4) \) to \( (0, 0) \). So the integral \( \int_{-3}^{0} f'(x) \, dx \) is the area between the graph of \( f' \), the \( x \)-axis, and the vertical lines \( x = -3 \) and \( x = 0 \). Since the graph is above the \( x \)-axis (from \( x = -3 \) to \( x = 0 \), \( f'(x) \) is positive), the integral is the area of the triangle with base \( 3 \) (from \( x = -3 \) to \( x = 0 \), the length is \( 0 - (-3) = 3 \)) and height \( 4 \) (at \( x = -3 \), \( f'(x) = 4 \), and at \( x = 0 \), \( f'(x) = 0 \)). Wait, the area of a triangle is \( \frac{1}{2} \times base \times height \). So base is \( 3 \) (from \( -3 \) to \( 0 \)), height is \( 4 \) (the \( y \)-value at \( x = -3 \)). So the area is \( \frac{1}{2} \times 3 \times 4 = 6 \). Wait, but let's check the intervals again. Wait, from \( x = -6 \) to \( x = -3 \), \( f'(x) = 4 \) (horizontal line). Then from \( x = -3 \) to \( x = 0 \), it's a line from \( (-3, 4) \) to \( (0, 0) \). Wait, no, the original graph: at \( x = -3 \), the point is \( (-3, 4) \), then at \( x = 0 \), it's \( (0, 0) \). So from \( x = -3 \) to \( x = 0 \), the graph is a line with slope \( (0 - 4)/(0 - (-3)) = -4/3 \), but the area under \( f'(x) \) from \( x = -3 \) to \( x = 0 \) is the integral of \( f'(x) \) from \( -3 \) to \( 0 \), which is the area of the region bounded by \( f'(x) \), the \( x \)-axis, \( x = -3 \), and \( x = 0 \). Since \( f'(x) \) is positive in this interval (from \( y = 4 \) at \( x = -3 \) to \( y = 0 \) at \( x = 0 \)), the integral is the area of the triangle with base \( 3 \) (from \( -3 \) to \( 0 \)) and height \( 4 \) (the maximum \( y \)-value in this interval). So the area is \( \frac{1}{2} \times 3 \times 4 = 6 \). Wait, but actually, from \( x = -3 \) to \( x = 0 \), the graph is a line from \( (-3, 4) \) to \( (0, 0) \), so the integral \( \int_{-3}^{0} f'(x) \, dx \) is the area of the triangle with vertices at \( (-3, 0) \), \( (0, 0) \), and \( (-3, 4) \). So that area is \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6 \).

Wait, but also, from \( x = -6 \) to \( x = -3 \), \( f'(x) = 4 \), which is a horizontal line. So from \( x = -6 \) to \( x = -3 \), the integral is \( 4 \times ( -3 - (-6) ) = 4 \times 3 = 12 \). From \( x = -9 \) to \( x = -6 \), it's a line from \( (-9, 0) \) to \( (-6, 4) \), so the area is \( \frac{1}{2} \times 3 \times 4 = 6 \). But we need the integral from \( x = -3 \) to \( x = 0 \). Wait, no, we need \( \int_{-3}^{0} f'(x) \, dx \). Let's re-express the problem. We have \( f(0) = 4 \), and we need to find \( f(-3) \). Using the Fundamental Theorem of Calculus: \( f(0) - f(-3) = \int_{-3}^{0} f'(x) \, dx \), so \( f(-3) = f(0) - \int_{-3}^{0} f'(x) \, dx \). Wait, no: \( f(b) - f(a) = \int_{a}^{b} f'(x) \, dx \). So if \( a = -3 \) and \( b = 0 \), then \( f(0) - f(-3) = \int_{-3}^{0} f'(x) \, dx \), so \( f(-3) = f(0) - \int_{-3}^{0} f'(x) \, dx \).

Now, let's compute \( \int_{-3}^{0} f'(x) \, dx \). The graph of \( f'(x) \) from \( x = -3 \) to \( x = 0 \) is a line from \( (-3, 4) \) to \( (0, 0) \). So this is a triangle with base length \( 0 - (-3) = 3 \) and height \( 4 \) (since at \( x = -3 \), \( f'(x) = 4 \), and at \( x = 0 \), \( f'(x) = 0 \)). The area of a triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \), so \( \int_{-3}^{0} f'(x) \, dx = \frac{1}{2} \times 3 \times 4 = 6 \).

Wait, but hold on, is the graph from \( x = -3 \) to \( x = 0 \) a triangle? Let's check the coordinates. At \( x = -3 \), the point is \( (-3, 4) \); at \( x = 0 \), the point is \( (0, 0) \). So the region under \( f'(x) \) from \( x = -3 \) to \( x = 0 \) is a triangle with vertices at \( (-3, 0) \), \( (0, 0) \), and \( (-3, 4) \). So the base is along the \( x \)-axis from \( -3 \) to \( 0 \), length \( 3 \), and the height is \( 4 \) (the vertical distance from \( y = 0 \) to \( y = 4 \) at \( x = -3 \)). So the area is indeed \( \frac{1}{2} \times 3 \times 4 = 6 \).

Now, since \( f(0) = 4 \), and \( f(0) - f(-3) = \int_{-3}^{0} f'(x) \, dx = 6 \), then \( 4 - f(-3) = 6 \), so \( -f(-3) = 6 - 4 = 2 \), so \( f(-3) = 4 - 6 = -2 \)? Wait, no, wait: \( f(0) - f(-3) = \int_{-3}^{0} f'(x) \, dx \). So \( 4 - f(-3) = 6 \), so \( -f(-3) = 6 - 4 = 2 \), so \( f(-3) = -2 \)? Wait, that can't be right. Wait, maybe I messed up the direction of the integral. Wait, the Fundamental Theorem of Calculus is \( f(b) - f(a) = \int_{a}^{b} f'(x) \, dx \). So if we want \( f(-3) \), we can write \( f(-3) - f(0) = \int_{0}^{-3} f'(x) \, dx \), because \( a = 0 \) and \( b = -3 \). Then \( \int_{0}^{-3} f'(x) \, dx = -\int_{-3}^{0} f'(x) \, dx \). So \( f(-3) = f(0) + \int_{0}^{-3} f'(x) \, dx = f(0) - \int_{-3}^{0} f'(x) \, dx \). So if \( \int_{-3}^{0} f'(x) \, dx = 6 \), then \( f(-3) = 4 - 6 = -2 \)? Wait, but let's re-examine the graph. Wait, from \( x = -3 \) to \( x = 0 \), the graph of \( f'(x) \) is a line from \( (-3, 4) \) to \( (0, 0) \), so the integral of \( f'(x) \) from \( -3 \) to \( 0 \) is the area under the curve, which is a triangle with base 3 and height 4, so area 6. But is the function positive or negative? From \( x = -3 \) to \( x = 0 \), \( f'(x) \) is positive (since it's above the \( x \)-axis), so the integral is positive 6. Then \( f(0) - f(-3) = 6 \), so \( 4 - f(-3) = 6 \), so \( f(-3) = 4 - 6 = -2 \). Wait, but let's check the graph again. Wait, maybe the graph from \( x = -3 \) to \( x = 0 \) is not a triangle. Wait, let's look at the coordinates again. The graph of \( f' \) at \( x = -3 \) is 4, at \( x = -2 \) is? Wait, from \( x = -3 \) to \( x = 0 \), the line goes from \( (-3, 4) \) to \( (0, 0) \), so the equation of the line is \( y = -\frac{4}{3}x \). Let's verify: when \( x = -3 \), \( y = -\frac{4}{3}(-3) = 4 \), correct. When \( x = 0 \), \( y = 0 \), correct. So the integral of \( y = -\frac{4}{3}x \) from \( x = -3 \) to \( x = 0 \) is \( \int_{-3}^{0} -\frac{4}{3}x \, dx \). Let's compute that: \( -\frac{4}{3} \times \frac{x^2}{2} \bigg|_{-3}^{0} = -\frac{2}{3}x^2 \bigg|_{-3}^{0} = -\frac{2}{3}(0 - 9) = -\frac{2}{3}(-9) = 6 \). So that's correct, the integral is 6. Then \( f(0) - f(-3) = 6 \), so \( 4 - f(-3) = 6 \), so \( f(-3) = 4 - 6 = -2 \). Wait, but let's check the other parts of the graph. Wait, maybe I missed a segment. Wait, from \( x = -6 \) to \( x = -3 \), the graph is horizontal at \( y = 4 \), so that's a rectangle with length \( -3 - (-6) = 3 \) and height 4, area 12. From \( x = -9 \) to \( x = -6 \), it's a triangle with base 3 and height 4, area 6. But we are only concerned with from \( x = -3 \) to \( x = 0 \). So the integral from \( -3 \) to \( 0 \) is 6. Then \( f(-3) = 4 - 6 = -2 \). Wait, but let's confirm with another approach. Suppose we consider the antiderivative. The integral of \( f'(x) \) from \( -3 \) to \( 0 \) is the net change in \( f(x) \) from \( x = -3 \) to \( x = 0 \). So \( f(0) - f(-3) = 6 \), so \( f(-3) = f(0) - 6 = 4 - 6 = -2 \). Yes, that seems correct.

Wait, but let's check the graph again. From \( x = -3 \) to \( x = 0 \), the graph of \( f' \) is a line from \( (-3, 4) \) to \( (0, 0) \), so the area under it is a triangle with base 3 (from -3 to 0 on the x-axis) and height 4 (the y-value at x=-3). The area of a triangle is (base * height)/2 = (3 * 4)/2 = 6. So the integral is 6. Then since \( f(0) = 4 \), and \( f(0) - f(-3) = 6 \), then \( f(-3) = 4 - 6 = -2 \).

Answer:

\( \boxed{-2} \)