Question

the function g is shown below. if f is the function defined by $f(x)=\int_{4}^{x}g(t)dt$, find the value of $f(-2)$ in simplest form.

Explanation:

Step1: Apply Fundamental Theorem of Calculus

By the Fundamental Theorem of Calculus, if \( f(x)=\int_{a}^{x}g(t)dt \), then \( f'(x) = g(x) \). So we need to find \( g(-2) \).

Step2: Analyze the graph at \( x = -2 \)

Looking at the graph of \( g \), at \( x=-2 \), the point is on the line segment. We can find the slope or directly read the value. From the graph, when \( x = -2 \), \( g(-2)=0 \)? Wait, no, wait. Wait, the graph: let's check the line. Wait, the graph has a segment from, let's see, the point at \( x=-2 \) is the vertex? Wait, no, looking at the graph, the line from \( x=-2 \) (where it crosses the x-axis) up to \( x = 3 \) or 4? Wait, no, the graph: at \( x=-2 \), the function \( g(t) \) at \( t=-2 \): let's see the graph. Wait, the graph has a part where from \( x=-10 \) to \( x=-4 \), it's a horizontal line at \( y=-6 \)? No, wait the black dots: one at \( x=-10, y=-6 \), another at \( x=-4, y=-6 \), then a line from \( x=-4 \) to \( x=-2 \) (where it crosses the x-axis), then a line from \( x=-2 \) to \( x = 3 \) (or 4) where \( y = 4 \), then horizontal. Wait, at \( x=-2 \), the function \( g(t) \) at \( t=-2 \): the line from \( x=-4 \) to \( x=-2 \): let's find the slope. From \( x=-4 \), what's the y-value? At \( x=-4 \), the point is \( (-4, -6) \)? Wait no, the black dot at \( x=-4 \) is \( (-4, -6) \)? Then at \( x=-2 \), it's \( (-2, 0) \). So the line from \( (-4, -6) \) to \( (-2, 0) \). The slope is \( \frac{0 - (-6)}{-2 - (-4)}=\frac{6}{2}=3 \). Wait, but at \( x=-2 \), the value of \( g(-2) \) is 0? Wait no, when \( x=-2 \), the graph crosses the x-axis, so \( g(-2)=0 \)? Wait, no, the Fundamental Theorem says \( f'(x)=g(x) \), so we need \( g(-2) \). Looking at the graph, at \( t=-2 \), the function \( g(t) \) is 0? Wait, no, let's re-examine. The graph: the line from \( x=-4 \) (black dot) to \( x=-2 \) (where it meets the x-axis) – so at \( x=-2 \), \( g(-2)=0 \)? Wait, no, maybe I misread. Wait, the problem is \( f(x)=\int_{4}^{x}g(t)dt \), so by Fundamental Theorem of Calculus, Part 1, \( f'(x)=g(x) \). So we need to find \( g(-2) \). Now, looking at the graph of \( g \), at \( x=-2 \), what's \( g(-2) \)? The graph has a segment from \( x=-4 \) (let's say the point is \( (-4, -6) \)) to \( x=-2 \) (where it's on the x-axis, so \( y=0 \)). So at \( x=-2 \), \( g(-2)=0 \)? Wait, no, that can't be. Wait, maybe the line from \( x=-2 \) to \( x=3 \) (or 4) is increasing. Wait, no, let's check the graph again. The graph: the part from \( x=-2 \) to \( x=3 \) (or 4) is a line with positive slope. Wait, no, the key is that \( f'(x)=g(x) \), so \( f'(-2)=g(-2) \). Now, looking at the graph, at \( x=-2 \), the function \( g(t) \) at \( t=-2 \): the graph crosses the x-axis at \( x=-2 \), so \( g(-2)=0 \)? Wait, no, maybe I made a mistake. Wait, the graph: the left part: from \( x=-10 \) to \( x=-4 \), it's horizontal at \( y=-6 \). Then from \( x=-4 \) to \( x=-2 \), it's a line going up to \( x=-2 \) (where \( y=0 \)). Then from \( x=-2 \) to \( x=3 \) (or 4), it's a line going up to \( y=4 \), then horizontal. So at \( x=-2 \), \( g(-2)=0 \)? Wait, no, when \( t=-2 \), the value of \( g(t) \) is 0? Then \( f'(-2)=g(-2)=0 \)? Wait, that can't be. Wait, no, maybe I misread the limits. The integral is from 4 to x, so \( f(x)=\int_{4}^{x}g(t)dt \). So by Fundamental Theorem of Calculus, \( f'(x)=g(x) \), regardless of the lower limit (as long as it's a constant). So we need \( g(-2) \). Now, looking at the graph, at \( t=-2 \), what's \( g(t) \)? Let's look at the graph again. The graph: at \( x=-2 \), the function is on the line that goes from \( x=-4 \) (where y is -6) to \( x=-2 \) (where y is 0). So at \( x=-2 \), \( g(-2)=0 \). Wait, but that seems odd. Wait, maybe the graph is different. Wait, the user's graph: the x-axis is from -10 to 10, y from -8 to 8. The graph has a horizontal segment at y=-6 from x=-10 to x=-4, then a line from x=-4 (y=-6) to x=-2 (y=0), then a line from x=-2 (y=0) to x=3 (y=4), then horizontal from x=3 to x=9 (y=4). So at x=-2, g(-2)=0. Therefore, f'(-2)=g(-2)=0? Wait, no, that can't be. Wait, maybe I messed up the direction. Wait, the integral is from 4 to x, so when x < 4, the integral is \( -\int_{x}^{4}g(t)dt \), but the Fundamental Theorem still applies: \( f'(x)=g(x) \), because the derivative of \( \int_{a}^{x}g(t)dt \) is \( g(x) \) for any a, by FTC Part 1. So regardless of a, as long as it's a constant, the derivative is g(x). So we need g(-2). Looking at the graph, at t=-2, g(t) is 0. So f'(-2)=0? Wait, no, wait the line from x=-2 to x=3: from x=-2 (y=0) to x=3 (y=4). The slope there is \( \frac{4 - 0}{3 - (-2)}=\frac{4}{5} \)? No, wait x=3, y=4? The black dot at x=3 is (3,4), then from x=3 to x=9, it's horizontal at y=4. So from x=-2 to x=3, the line is from (-2, 0) to (3, 4), slope \( \frac{4}{5} \), but at x=-2, the value is 0. So g(-2)=0. Therefore, f'(-2)=0? Wait, but that seems wrong. Wait, maybe the graph at x=-2 is not 0. Wait, let's check the coordinates again. The point at x=-4: is that (-4, -6)? The black dot at x=-4: yes, then at x=-2, it's (-2, 0). So the line from (-4, -6) to (-2, 0): when t=-2, g(t)=0. So f'(-2)=g(-2)=0.

Wait, but maybe I made a mistake. Let's re-express:

FTC Part 1: If \( f(x)=\int_{a}^{x}g(t)dt \), then \( f'(x)=g(x) \), provided g is continuous on an interval containing a and x.

So here, a=4, x=-2. We need to check if g is continuous at t=-2 (and on the interval from -2 to 4, but since we just need the derivative at x=-2, we need g(-2).

From the graph, at t=-2, g(t)=0. So f'(-2)=g(-2)=0.

Answer:

\( \boxed{0} \)