a function $f(x)$ has a given value $f(0)= - 3$. a graph of the derivative $f(x)$ is given below. what is…

Explanation:

Step1: Apply the Net Change Theorem

The Net Change Theorem states that $F(b) - F(a) = \int_{a}^{b} F'(x) dx$. We need to find $F(2)$, given $F(0) = -3$. Thus, $F(2) = F(0) + \int_{0}^{2} F'(x) dx$.

Step2: Determine the equation of $F'(x)$

From the graph, $F'(x)$ is a straight line. We can identify two points on the line: $(0, -1)$ and $(1, 2)$. The slope $m$ is $m = \frac{2 - (-1)}{1 - 0} = \frac{3}{1} = 3$. The y-intercept is $c = -1$ (the value of $F'(x)$ when $x=0$). So, the equation of the derivative is $F'(x) = 3x - 1$.

Step3: Calculate the definite integral $\int_{0}^{2} F'(x) dx$

We integrate $F'(x)$ from $0$ to $2$: $$ \int_{0}^{2} (3x - 1) dx $$ $$ = \left[ \frac{3x^2}{2} - x \right]_{0}^{2} $$ $$ = \left( \frac{3(2)^2}{2} - 2 \right) - \left( \frac{3(0)^2}{2} - 0 \right) $$ $$ = \left( \frac{3 \cdot 4}{2} - 2 \right) - (0) $$ $$ = (6 - 2) - 0 $$ $$ = 4 $$ Alternatively, we can calculate the area geometrically. The graph of $F'(x)$ from $x=0$ to $x=2$ forms two triangles with the x-axis. The x-intercept of $F'(x)=3x-1$ is at $x = \frac{1}{3}$. Area of the triangle below the x-axis (from $x=0$ to $x=\frac{1}{3}$): $A_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{1}{3} \times |-1| = \frac{1}{6}$. The integral value is $-\frac{1}{6}$. Area of the triangle above the x-axis (from $x=\frac{1}{3}$ to $x=2$): $A_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2 - \frac{1}{3}) \times (3(2)-1) = \frac{1}{2} \times \frac{5}{3} \times 5 = \frac{25}{6}$. The integral value is $\frac{25}{6}$. The definite integral is $-\frac{1}{6} + \frac{25}{6} = \frac{24}{6} = 4$.

Step4: Calculate $F(2)$

Using the formula from Step 1: $F(2) = F(0) + \int_{0}^{2} F'(x) dx$. Given $F(0) = -3$ and we found $\int_{0}^{2} F'(x) dx = 4$. $$ F(2) = -3 + 4 $$ $$ F(2) = 1 $$

Answer:

$1$