High School

8.3 Given: f(x) = -6x^3

8.3.1 Determine f'(x) from first principles.

8.3.2 Write down how you will restrict the domain of f such that f^{-1}, the inverse of f, is a function.

8.3.3 Determine the equation of f^{-1} for f^{-1}(x) ≤ 0. Write your answer in the form y = ...

Answer :

Let's approach the problem step by step:

8.3.1 Determine [tex]f'(x)[/tex] from first principles.

To find the derivative [tex]f'(x)[/tex] from first principles, also known as using the definition of the derivative, we start with:

[tex]f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}[/tex]

Given [tex]f(x) = -6x^{3}[/tex], let's calculate [tex]f(x+h)[/tex]:

[tex]f(x+h) = -6(x+h)^3[/tex]

Expanding [tex](x+h)^3[/tex] using the binomial theorem gives:

[tex](x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3[/tex]

Thus:

[tex]f(x+h) = -6(x^3 + 3x^2h + 3xh^2 + h^3) = -6x^3 - 18x^2h - 18xh^2 - 6h^3[/tex]

Now, substitute back into the derivative formula:

[tex]f'(x) = \lim_{h \to 0} \frac{(-6x^3 - 18x^2h - 18xh^2 - 6h^3) - (-6x^3)}{h}[/tex]

This simplifies to:

[tex]f'(x) = \lim_{h \to 0} \frac{-18x^2h - 18xh^2 - 6h^3}{h}[/tex]

Factor out [tex]h[/tex] from the numerator:

[tex]f'(x) = \lim_{h \to 0} (-18x^2 - 18xh - 6h^2)[/tex]

When [tex]h \to 0[/tex], the expression simplifies to:

[tex]f'(x) = -18x^2[/tex]

8.3.2 Write down how you will restrict the domain of [tex]f[/tex] such that [tex]f^{-1}[/tex], the inverse of [tex]f[/tex], is a function.

The function [tex]f(x) = -6x^3[/tex] is a cubic function, and it is not one-to-one over the entire set of real numbers. However, for [tex]f^{-1}[/tex] to be a function, [tex]f[/tex] must be one-to-one, which means we need to restrict its domain.

One common way to make a cubic function one-to-one is to restrict its domain to either [tex]x \geq 0[/tex] or [tex]x \leq 0[/tex]. Let's restrict the domain to [tex]x \geq 0[/tex].

8.3.3 Determine the equation of [tex]f^{-1}[/tex] for [tex]f^{-1}(x) \leq 0[/tex]. Write your answer in the form [tex]y = \ldots[/tex]

Given [tex]f(x) = -6x^3[/tex] with restricted domain [tex]x \geq 0[/tex], we need to solve [tex]y = -6x^3[/tex] for [tex]x[/tex]:

  1. Solve for [tex]x[/tex] in terms of [tex]y[/tex]:
    [tex]y = -6x^3[/tex]
    Divide both sides by [tex]-6[/tex]:
    [tex]\frac{y}{-6} = x^3[/tex]
    Take the cube root of both sides:
    [tex]x = \sqrt[3]{\frac{y}{-6}}[/tex]

    Since we want [tex]f^{-1}(x) \leq 0[/tex], this implies:

    [

y \leq 0
]

Thus, the equation of [tex]f^{-1}(x)[/tex] is:

[tex]y = \sqrt[3]{\frac{x}{-6}}[/tex]

for [tex]x \leq 0[/tex]. This is the equation of the inverse function, considering the restricted domain.

Other Questions