Hard calculus question

Question

g(x) = f(x) . tanx+ kx, where k is a real number

f is differentiable for all x

f(pi/4) = 4

f'(pi/4) = -2

For what values of x, if any, in the interval 0 < x < 2x will the derivative of g fail to exist. Justify your answer.

If g'(pi/4) = 6, find the value of k

Answer 1

g(x) is not defined where tan(x) is not defined, which is

for x = π/2 + kπ, for any integer k.

For other values of x

g'(x) = f'(x)tax(x) + f(x)/cos²(x) + k (1)

So again g'(x) is not defined where tan(x) is not defined, or where cos(x) = 0.

That is for x = π/2 + kπ for any integer k.

Assuming a typo in your question, and that you are loking for values of x

in the interval 0 < x < 2π, where g'(x) does not exist, then the answer is

x = π/2 and 3π/2.

To find the value of k, into (1) substitute given values for

f(π/4), f'(π/4), g'(π/4):

g'(π/4) = f(π/4)tan(π/4) + f(π/4)/cos²(π/4) + k

6 = (-2)×1 + 4/(1/2) + k

k = 0

Answer 2

Do you mean g(x)= f(x) tan x + kx or g(x)= f(x) [tanx+kx]. Either way we can use the product rule to proceed because you can easily find the value of tan x and its derivative for x=π/4 (a special angle).

The part where g(x) is undefined occurs when tan x is undefined or when x = π/2 or 3π/2. I believe you mean the region x∈ (0,2π).

Let me know if you need to go through the steps.