a function $f(x)$ has a given value $f(0)= - 2$. a graph of the derivative $f(x)$ is given below. what is…

Explanation:

Step1: Apply the Fundamental Theorem of Calculus.

The Fundamental Theorem of Calculus states that if $F(x)$ is an antiderivative of $F'(x)$, then $\int_{a}^{b} F'(x) dx = F(b) - F(a)$. We want to find $F(1)$, and we are given $F(0) = -2$. So, we can write: $F(1) - F(0) = \int_{0}^{1} F'(x) dx$ $F(1) = F(0) + \int_{0}^{1} F'(x) dx$

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

The graph of $F'(x)$ is a straight line. We can identify two points on the line: $(0, -1)$ and $(1, 1)$. The slope $m$ of the line is $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-1)}{1 - 0} = \frac{2}{1} = 2$. The y-intercept $c$ is the value of $F'(x)$ when $x=0$, which is $-1$. So, the equation of the line is $F'(x) = 2x - 1$.

Step3: Calculate the definite integral $\int_{0}^{1} F'(x) dx$.

We need to evaluate the integral of $F'(x) = 2x - 1$ from $0$ to $1$: $$ \int_{0}^{1} (2x - 1) dx $$ $$ = \left[ \frac{2x^2}{2} - x \right]{0}^{1} $$ $$ = [x^2 - x]{0}^{1} $$ $$ = (1^2 - 1) - (0^2 - 0) $$ $$ = (1 - 1) - 0 $$ $$ = 0 - 0 = 0 $$ Alternatively, we can calculate the area under the graph of $F'(x)$ from $x=0$ to $x=1$. The graph forms two triangles: one below the x-axis from $x=0$ to $x=0.5$ (where $2x-1=0$) with base $0.5$ and height $-1$, and one above the x-axis from $x=0.5$ to $x=1$ with base $0.5$ and height $1$. Area = $\frac{1}{2} \times 0.5 \times (-1) + \frac{1}{2} \times 0.5 \times 1 = -0.25 + 0.25 = 0$.

Step4: Calculate $F(1)$.

Using the formula from Step1: $F(1) = F(0) + \int_{0}^{1} F'(x) dx$ Given $F(0) = -2$ and we calculated $\int_{0}^{1} F'(x) dx = 0$. $F(1) = -2 + 0$ $F(1) = -2$

Answer:

$F(1) = -2$