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Rzqust [24]
3 years ago
5

Find the inverse of the function. f(x) = 6x3 - 8

Mathematics
2 answers:
mr Goodwill [35]3 years ago
8 0
Replace <span><span>f<span>(x)</span></span><span>fx</span></span> with <span>yy</span>.<span><span>y=<span><span>6<span>x3</span></span><span>−8</span></span></span><span>y=<span><span>6<span>x3</span></span><span>-8</span></span></span></span>Interchange the variables.<span><span>x=<span><span>6<span>y3</span></span><span>−8</span></span></span><span>x=<span><span>6<span>y3</span></span><span>-8</span></span></span></span>Solve for <span>yy</span>.
<span><span>y=<span><span>3<span>√<span>36<span>(<span>x+8</span>)</span></span></span></span>6</span></span><span>y=<span><span><span>36<span>x+8</span></span>3</span>6</span></span></span>Solve for <span>yy</span> and replace with <span><span><span>f<span>−1</span></span><span>(x)</span></span><span><span>f<span>-1</span></span>x</span></span>.
<span><span><span>f<span>−1</span></span><span>(x)</span></span>=<span><span>3<span>√<span>36<span>(<span>x+8</span>)</span></span></span></span><span>6</span></span></span>
Softa [21]3 years ago
6 0

Answer:

f^{-1}(x) = \sqrt[3]{\frac{x+8}{6}}

Step-by-step explanation:

Given function,

f(x)=6x^3-8

Replace f(x) by y,

y=6x^3-8

Switch x and y,

x=6y^3-8

Isolate y in the left side,

-6y^3=-x-8

6y^3=x+8

y^3=\frac{x+8}{6}

y=\sqrt[3]{\frac{x+8}{6}}

Replace y by f^{-1}(x)

\implies f^{-1}(x) = \sqrt[3]{\frac{x+8}{6}}

Which is the required inverse of the given function.

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See attachment for the graph of the hyperbola 12x^2 - 3y^2 - 108 = 0

<h3>How to graph the hyperbola?</h3>

The equation of the hyperbola is given as:

12x^2 - 3y^2 - 108 = 0

Start by calculating the transverse axis

So, we have:

<u>Transverse axis</u>

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

Where

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The transverse axis is calculated as:

y = ±b/a(x - h) + k

So, we have:

y = ±6/3(x - 0) + 0

Evaluate the difference and sum

y = ±6/3x

Evaluate the quotient

y = ±2x

This means that the transverse axes are y = 2x and y =-2x

<u>The vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

<u>The co-vertices</u>

In the above section, we have:

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This means that

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See attachment for the graph of the hyperbola

Read more about hyperbola at:

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