Find the Tangent Line to g(x) at x=pi using Chain Rule | Math Homework Solution

[Speedking96]

Homework Statement

Let g(x) = f(sin(2x) f(cos x)), where f(0) = 2, f'(0) = 3, f(-1) = -1/3 , and f'(-1) = -1. Find the equation of the tangent line to the curve of y = g(x) at x = pi.

2. The attempt at a solution

Point of Tangent: (pi, 2)
g(pi) = f(sin(2pi) f(cos pi)) = f(0 * f(-1)) = f(0) = 2

g(x) = f(sin(2x) f(cos x))
= f'(sin(2x)* f(cos x)) * ([2cos(x) f(cos x)] + [f'(cos x) * (-sin x) * (sin 2x])
= f'(sin(2pi) * f(cos pi)) *([2cos(pi) f(cos pi)] + [f'(cos pi) * (-sin pi) * (sin 2pi])
= f'(0* f(-1)) *([(-2) f(-1)] + [f'(-1) * (0) * (0)])
= f'(0) *(-2) f(-1)
= (3)(-2)(-1/3) = 2

Tangent line:

y = 2x + b
2 = 2(pi) + b
b = 2 - (2)(pi)

y = 2x + 2 - 2pi

 

[LCKurtz]

Homework Statement

Let g(x) = f(sin(2x) f(cos x)), where f(0) = 2, f'(0) = 3, f(-1) = -1/3 , and f'(-1) = -1. Find the equation of the tangent line to the curve of y = g(x) at x = pi.

2. The attempt at a solution

Point of Tangent: (pi, 2)
g(pi) = f(sin(2pi) f(cos pi)) = f(0 )* f(-1)) = f(0) = 2

Isn't f(-1) = -1/3?

g(x) = f(sin(2x) f(cos x))
g'(x) = f'(sin(2x)* f(cos x)) * ([2cos(x) f(cos x)] + [f'(cos x) * (-sin x) * (sin 2x])

You need to fix the derivative.

 

[Speedking96]

Isn't f(-1) = -1/3?

Isn't f(-1) = -1/3?

You need to fix the derivative.

g(pi) = f[ sin(2pi) * f (cos pi) ] = f [ 0 * -1/3 ] = f(0) = 2

What's wrong with the derivative? I applied the chain rule.

 

[Doc Al]

g(x) = f(sin(2x) f(cos x))
= f'(sin(2x)* f(cos x)) * ([2cos(x) f(cos x)] + [f'(cos x) * (-sin x) * (sin 2x])

Looks like you messed up the derivative of sin(2x).

 

[Speedking96]

Oh! It should be 2cos(2x). Thank you!