current attempt in progress\nthe graph of a d...
current attempt in progress\nthe graph of a derivative f(x) is shown in the figure below.\nfill in the table of values for f(x) given that f(0) = 9.\nenter the exact answers.\n| x | 0 | 1 | 2 | 3 | 4 | 5 | 6 |\n| f(x) | 9 | i | i | i | i | i | i |
Answer
# Explanation:
## Step1: Recall the fundamental theorem of calculus
The change in $f(x)$ from $x = a$ to $x = b$ is given by $\int_{a}^{b}f^{\prime}(x)dx=f(b)-f(a)$. So $f(x)=f(0)+\int_{0}^{x}f^{\prime}(t)dt$.
## Step2: Calculate $f(1)$
The integral $\int_{0}^{1}f^{\prime}(x)dx$ is the area under the curve $y = f^{\prime}(x)$ from $x = 0$ to $x = 1$. Since $f^{\prime}(x)= - 7$ on $[0,1]$, then $\int_{0}^{1}f^{\prime}(x)dx=-7\times1=-7$. So $f(1)=f(0)+\int_{0}^{1}f^{\prime}(x)dx=9 - 7=2$.
## Step3: Calculate $f(2)$
$\int_{0}^{2}f^{\prime}(x)dx=\int_{0}^{1}f^{\prime}(x)dx+\int_{1}^{2}f^{\prime}(x)dx$. Since $f^{\prime}(x)=-7$ on $[0,1]$ and $[1,2]$, $\int_{0}^{2}f^{\prime}(x)dx=-7\times2=-14$. So $f(2)=f(0)+\int_{0}^{2}f^{\prime}(x)dx=9-14 = - 5$.
## Step4: Calculate $f(3)$
$\int_{0}^{3}f^{\prime}(x)dx=\int_{0}^{1}f^{\prime}(x)dx+\int_{1}^{2}f^{\prime}(x)dx+\int_{2}^{3}f^{\prime}(x)dx$. Since $f^{\prime}(x)=-7$ on $[0,1]$, $[1,2]$ and $[2,3]$, $\int_{0}^{3}f^{\prime}(x)dx=-7\times3=-21$. So $f(3)=f(0)+\int_{0}^{3}f^{\prime}(x)dx=9-21=-12$.
## Step5: Calculate $f(4)$
$\int_{0}^{4}f^{\prime}(x)dx=\int_{0}^{3}f^{\prime}(x)dx+\int_{3}^{4}f^{\prime}(x)dx$. The integral $\int_{3}^{4}f^{\prime}(x)dx$ is the area of a trapezoid with bases $b_1=-7$ and $b_2 = 0$ and height $h = 1$. The area of the trapezoid $A=\frac{(-7 + 0)\times1}{2}=-\frac{7}{2}$. $\int_{0}^{4}f^{\prime}(x)dx=-21-\frac{7}{2}=-\frac{42 + 7}{2}=-\frac{49}{2}$. So $f(4)=f(0)+\int_{0}^{4}f^{\prime}(x)dx=9-\frac{49}{2}=\frac{18 - 49}{2}=-\frac{31}{2}$.
## Step6: Calculate $f(5)$
$\int_{4}^{5}f^{\prime}(x)dx$ is the area of a triangle with base $b = 1$ and height $h=7$. So $\int_{4}^{5}f^{\prime}(x)dx=\frac{1\times7}{2}=\frac{7}{2}$. $\int_{0}^{5}f^{\prime}(x)dx=\int_{0}^{4}f^{\prime}(x)dx+\int_{4}^{5}f^{\prime}(x)dx=-\frac{49}{2}+\frac{7}{2}=-\frac{42}{2}=-21$. So $f(5)=f(0)+\int_{0}^{5}f^{\prime}(x)dx=9-21=-12$.
## Step7: Calculate $f(6)$
$\int_{5}^{6}f^{\prime}(x)dx=7\times1 = 7$. $\int_{0}^{6}f^{\prime}(x)dx=\int_{0}^{5}f^{\prime}(x)dx+\int_{5}^{6}f^{\prime}(x)dx=-21 + 7=-14$. So $f(6)=f(0)+\int_{0}^{6}f^{\prime}(x)dx=9-14=-5$.
# Answer:
| $x$ | $f(x)$ |
| ---- | ---- |
| 0 | 9 |
| 1 | 2 |
| 2 | - 5 |
| 3 | - 12 |
| 4 | $-\frac{31}{2}$ |
| 5 | - 12 |
| 6 | - 5 |