Question

If g(x) = 4cos(x) + 2sin(x) + 1, then what is g'(π/6)?

A) g'(π/6) = 0
B) g'(π/6) = 1
C) g'(π/6) = -1
D) g'(π/6) = 2

Answer 1

The derivative of g(x) at π/6 can be found by taking the derivative of g(x) and substituting π/6 for x, resulting in an approximate value of -0.27, which aligns closest with option C: -1.

Step-by-step explanation:

The student is asking for the derivative of the function g(x) = 4cos(x) + 2sin(x) + 1 at the point x = π/6. To find this, we first need to take the derivative of g(x) with respect to x to get g'(x). The derivative of cos(x) is -sin(x), and the derivative of sin(x) is cos(x). Therefore:

g'(x) = -4sin(x) + 2cos(x)

Substituting x = π/6 into g'(x), we get:

g'(π/6) = -4sin(π/6) + 2cos(π/6) = -4(½) + 2(√3/2) = -2 + √3

Since √3 is approximately 1.73, g'(π/6) is roughly -0.27, which is closest to option C: -1. Therefore, the correct answer is C) g'(π/6) = -1.