AP+Review+10+Problem+1

Jacob Chess Period 3

f(x) = integral((4+t^2)^(1/2))dt on [0,3x]

g(x) = f(sin(x))

a) Find f ' (x) and g ' (x).

f ' (x) = (((4+(3x)^2)^(1/2)) x 3) - ((4+(0)^2)^(1/2)) f ' (x) = 3(4+9x^2)^(1/2)

g ' (x) = (((4+(3sin(x))^2)^(1/2)) x 3cos(x)) g ' (x) = (3cos(x))(4+9(sin(x))^2)

f ' (x) is 3(4+9x^2)^(1/2) and g ' (x) is (3cos(x))(4+9(sin(x))^2).

b) Write an equation for the line tangent to the graph of y = g(x) at x = pi.

You need a point and a slope:

g(x) = integral((4+t^2)^(1/2))dt on [0,(3sin(x))] g(pi) = integral((4+t^2)^(1/2))dt on [0,(3sin(pi))] g(pi) = integral((4+t^2)^(1/2))dt on [0,0] g(pi) = 0

Point: (pi,0)

g ' (pi) = (3cos(pi))(4+9(sin(pi))^2) g ' (pi) = -3(4)^(1/2) g ' (pi) = -6

Slope: -6

y - 0 = -6(x-pi) y = -6x + 6(pi)

The line y = -6x + 6(pi) is tangent to g(x) at x = pi.

c) Write, but do not evaluate, an integral expression that represents the maximum value of g on the interval [0,pi]. Justify your answer.

integral((4+t^2)^(1/2))dt on [0,(3sin(x))] represents the maximum value of g on the interval [0,pi].