find the eq to the tangent line to the curve y=x^3 g(x) at x=1 if g(1)=2 f(1)=3 f'(1)=1 g'(1)=-2

i cant understand how the answer is y=4x-2

you are probably not getting a response to your question since others along with me cannot figure out what you mean by

.... g(x) at x=1 if g(1)=2 f(1)=3 f'(1)=1 g'(1)=-2

if y = 4x-2 is the tangent equation then the contact point would have an x value of 2/√3, and I see no √3 anywhere.

you speak of g(x) and f(x) but state your equation y = x^3.
Quite confusing.

To find the equation of the tangent line to the curve y = g(x) at x = 1, we first need to find the slope of the tangent line. The slope of the tangent line is equal to the derivative of the function g(x) evaluated at x = 1 (g'(1)).

Given that g'(1) = -2, we know that the slope of the tangent line is -2.

Next, to find the equation of the tangent line, we need a point on the line. Since the point x = 1 lies on both the curve y = g(x) and the tangent line, we can use this point to determine the equation.

Given that g(1) = 2, we know that when x = 1, y = 2. Therefore, we have the point (1, 2) on the tangent line.

Using the slope-intercept form of a linear equation, which is y = mx + b (m being the slope and b being the y-intercept), we can substitute the values we have to find the y-intercept (b).

Using the point (1, 2) and the slope (-2), we have the equation:

2 = -2(1) + b

Simplifying, we get:

2 = -2 + b

Adding 2 to both sides, we get:

4 = b

Now, we have both the slope (-2) and the y-intercept (b = 4), so we can write the equation of the tangent line:

y = -2x + 4

However, you mentioned in your question that the answer is y = 4x - 2. It seems there may have been a mistake in either the given values or the calculations. Double-check your given values and calculations to make sure you haven't missed anything.