Question

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!

Answer 1

Answer:

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 = $\dfrac{x^3 -3\cdot x^2 + 3 \cdot x - 1}{27}$

The slope is therefore;

$\dfrac{\mathrm{d} g(x)^{3}}{\mathrm{d} x} = \dfrac{27 \cdot (3 \cdot x^2 -6\cdot x +3 )}{729}$

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

$Slope \ of \ tangent \ of \ g(x)^3= -\dfrac{729}{27 \cdot (3 \cdot x^2 -6\cdot x +3 )} = -\dfrac{9}{x^2 -2\cdot x + 1}$

The value of the slope at x = 4 is  $-\dfrac{9}{4^2 -2\cdot 4 + 1} = \dfrac{-9}{9} = -1$

Therefore, we have;

y at x = 4

$y = \dfrac{4^3 -3\cdot 4^2 + 3 \cdot 4 - 1}{27} = \dfrac{27}{27} = 1$

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.