Consider the function f(x)= 1/2 cos(2x)

Question

1. Consider the function f(x)= 1/2 cos(2x)

(a) Compute ( f’(x)

(b) Find an equation for the tangent line to the graph of f(x) at x= π/3

(c) Compute f⁽¹¹⁾(x).

Answer 1

f '(x) = 1/2(-sin(2x)•(2) = -sin(2x)

For x = π/3, f(π/3) = 1/2cos(2•(π/3)) = 1/2cos(2π/3) = 12(-1/2) = -1/4

So the tangent point is (π/3, -1/4)

Calculate slope of the tangent line: At x = π/3 then f'(π/3) = -sin(2•π/3) = -sin(2π/3) = -√3/2

Use the point-slope form to get the equation of the tangent line:

y - y₁ = m(x - x₁) where m = -√3/2 and (x₁, y₁) = (π/3, -1/4)

y - -1/4 = -√3/2(x - π/3)

y + 1/4 = -√3/2(x - π/3)

To compute f⁽¹¹⁾ look for the pattern:

f(x) = 1/2cos(2x)

f⁽¹⁾ = -sin(2x)

f⁽²⁾ = -2cos(2x)

f⁽³⁾ = 4sin(2x)

f⁽⁴⁾ = 8cos(2x)

f⁽⁵⁾ = -16sin(2x)

f⁽⁶⁾ = -32cos(2x)

f⁽⁷⁾ = 64sin(2x)

f⁽⁸⁾ = 128cos(2x)

f⁽⁹⁾ = -256sin(2x)

f⁽¹⁰⁾ = -512cos(2x)

f⁽¹¹⁾ = 1024sin(2x)

Answer 2

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