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 = [tex]\dfrac{x^3 -3\cdot x^2 + 3 \cdot x - 1}{27}[/tex]
The slope is therefore;
[tex]\dfrac{\mathrm{d} g(x)^{3}}{\mathrm{d} x} = \dfrac{27 \cdot (3 \cdot x^2 -6\cdot x +3 )}{729}[/tex]
The slope of the tangent is the negative reciprocal of the slope of the line which gives;
[tex]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}[/tex]
The value of the slope at x = 4 is [tex]-\dfrac{9}{4^2 -2\cdot 4 + 1} = \dfrac{-9}{9} = -1[/tex]
Therefore, we have;
y at x = 4
[tex]y = \dfrac{4^3 -3\cdot 4^2 + 3 \cdot 4 - 1}{27} = \dfrac{27}{27} = 1[/tex]
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.