Consider the graph of f shown to the right. x y = f(t) a. Estimate the zero(s) of the area function A(x) = |f(t) dt for 0 s x = 10. 6 10 b. Estimate the points (if any) at which A has a local maximum or minimum. c. Sketch a rough graph of A for 0 x 10 without a scale on the y-axis. a. The zero(s) of A is/are approximately x =]. (Use a comma to separate answers as needed. Round to one decimal place as needed.) b. The area function A has a local maximum or minimum at approximately x =]. (Use a comma to separate answers as needed. Round to one decimal place as needed.) c. Choose the correct sketch below. O A. O B. O C. O D. Ay Ay Ay X X X 5 10 10 O- 5 O- 10 10x Let f(t) = 4t - 16 and consider the two area functions A(x) = f(t) dt and F(x) = f(t) dt. Complete parts (a) (c). a. Evaluate A(5) and A(6). Then use geometry to find an expression for A(x) for all x 2 4. The value of A(5) is . (Simplify your answer.) The value of A(6) is . (Simplify your answer.) Use geometry to find an expression for A(x) when x = 4. Type an expression using x as the variable.) b. Evaluate F(8) and F(9). Then use geometry to find an expression for F(x) for all x =7. The value of F(8) is . (Simplify your answer.) The value of F(9) is . (Simplify your answer.) Use geometry to find an expression for F(x) when x 2 7. (Type an expression using x as the variable.)c. Show that A(x) - F(x) is a constant and that A'(x) = F'(x) = f(x). First prove A(x) - F(x) is a constant. Given A(x) = 2x- - 16x + 32 and F(x) = 2x- - 16x + 14, subtract F(x) from A(X). The value of A(x) - F(x) is . (Simplify your answer.) Now prove that A'(x) = F'(x) = f(x). Given A(x) and F(x), take the derivative of A(x) and F(x) respectively and compare the results. A' (x) = - (2x2 - 16x + 32) = (Simplify your answer. Do not factor.) d F' (X) = - dx (2x2 - 16x+ 14) =(Simplify your answer. Do not factor.) Recall that f(t) = 4t - 16, substitute x for t to get f(x). f(x) =(Simplify your answer. Do not factor.) Thus, A'(x) = F'(x) = f(x)