The tangent line to the graph of y=g(x) at x=4 has equation y=-3x+11. What is the equation of the tangent line to the graph of y=g(x)^3 at x=4? Need correct answer and explanation as soon as possible! Will give brainliest!

The equation of the tangent of g(x)^3 at x = 4 is y = 3 - x

Explanation:

The tangent of y = g(x) = -3·x + 11

Therefore, the slope of g(x) = 1/3

The value of y = -3*4 + 11 = -1

The equation of the line g(x) is given as follows;

y - 1 = 1/3*(x - 4)

y - 1 = 1/3x - 4/3

y = 1/3x - 4/3 + 1 = 1/3x - 1/3

g(x) = 1/3x - 1/3

g(x)^3 = (1/3x - 1/3)^3 =

The slope is therefore;

The slope of the tangent is the negative reciprocal of the slope of the line which gives;

The value of the slope at x = 4 is  

Therefore, we have;

y at x = 4

Therefore, the equation of the tangent is given as follows;

y - 1 =(-1) × (x - 4) = 4 - x

y = 4 - 1 - x = 3 - x

The equation of the tangent of g(x)^3 at x = 4 is y = 3 - x.

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