Question

nice job! now answer the questions below.
answer attempt 1 out of 2
$f(6) = \square$ $f(7) = \square$ $f(7) - f(6) = \square$ $\int_{6}^{7} f(x) dx = \square$

Explanation:

To solve the problem, we analyze the graph of \( f(x) \) and use the Fundamental Theorem of Calculus for the integral of \( f'(x) \).

Step 1: Find \( f(6) \)

From the graph of \( f(x) \), we identify the \( y \)-value at \( x = 6 \). Looking at the graph, \( f(6) = -4 \).

Step 2: Find \( f(7) \)

From the graph of \( f(x) \), we identify the \( y \)-value at \( x = 7 \). Looking at the graph, \( f(7) = -3 \).

Step 3: Calculate \( f(7) - f(6) \)

Substitute the values of \( f(7) \) and \( f(6) \):
\[
f(7) - f(6) = -3 - (-4) = -3 + 4 = 1
\]

Step 4: Evaluate \( \int_{6}^{7} f'(x) \, dx \)

By the Fundamental Theorem of Calculus, \( \int_{a}^{b} f'(x) \, dx = f(b) - f(a) \). Here, \( a = 6 \) and \( b = 7 \), so:
\[
\int_{6}^{7} f'(x) \, dx = f(7) - f(6) = 1
\]

Final Answers:

• \( f(6) = \boxed{-4} \)
• \( f(7) = \boxed{-3} \)
• \( f(7) - f(6) = \boxed{1} \)
• \( \int_{6}^{7} f'(x) \, dx = \boxed{1} \)

Answer:

To solve the problem, we analyze the graph of \( f(x) \) and use the Fundamental Theorem of Calculus for the integral of \( f'(x) \).

Step 1: Find \( f(6) \)

From the graph of \( f(x) \), we identify the \( y \)-value at \( x = 6 \). Looking at the graph, \( f(6) = -4 \).

Step 2: Find \( f(7) \)

From the graph of \( f(x) \), we identify the \( y \)-value at \( x = 7 \). Looking at the graph, \( f(7) = -3 \).

Step 3: Calculate \( f(7) - f(6) \)

Substitute the values of \( f(7) \) and \( f(6) \):
\[
f(7) - f(6) = -3 - (-4) = -3 + 4 = 1
\]

Step 4: Evaluate \( \int_{6}^{7} f'(x) \, dx \)

By the Fundamental Theorem of Calculus, \( \int_{a}^{b} f'(x) \, dx = f(b) - f(a) \). Here, \( a = 6 \) and \( b = 7 \), so:
\[
\int_{6}^{7} f'(x) \, dx = f(7) - f(6) = 1
\]

Final Answers:

• \( f(6) = \boxed{-4} \)
• \( f(7) = \boxed{-3} \)
• \( f(7) - f(6) = \boxed{1} \)
• \( \int_{6}^{7} f'(x) \, dx = \boxed{1} \)