Solving functions algebraically (cube roots)

In summary: I do to it.I feel a bit stupid for not being able to solve this because the solving is not supposed to be the challenge, the concept of what makes the functions of inverses is.
  • #1
KatieLynn
65
0

Homework Statement


Show f and g are inverse functions or state that they are not.
f(x)= cube root of -8x-6 g(x)= -(x^3+6)/(8)

Homework Equations


You find inverses by plugging the equations into each other, if they are inverses then once you simplify the composed equation, it will equal x.

The Attempt at a Solution



cube root of {-8[-(x^3+6)/(8)]-6} I plugged g(x) into f(x) now I need to solve it, this is what I did but its not right because the answer should come out to be just x.

I multiplied everything by 8 to get it out of the denominator and simplified the numerator which left me with

cube root of {64x^3-432}

then I thought maybe you could pull out the 64 and x^3 from the cube root to get

4x times the cube root of {-432}

which is definitely not going to = x not matter what I do to it.

I feel a bit stupid for not being able to solve this because the solving is not supposed to be the challenge, the concept of what makes the functions of inverses is. I think I've just forgotten how to solve harder equations.
 
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  • #2
I think its just a case of a mistake in your algebra. Can you show your full working perhaps?
 
  • #3
Say f(x) = y = cube root of -8x-6.

Solve for [EDIT: x].

Hint: y^3 = -8x-6.

Replace x with y, and y with x. Check to see y = g(x).
 
Last edited:
  • #4
KatieLynn said:

Homework Statement


Show f and g are inverse functions or state that they are not.
f(x)= cube root of -8x-6 g(x)= -(x^3+6)/(8)

Homework Equations


You find inverses by plugging the equations into each other, if they are inverses then once you simplify the composed equation, it will equal x.


The Attempt at a Solution



cube root of {-8[-(x^3+6)/(8)]-6} I plugged g(x) into f(x) now I need to solve it, this is what I did but its not right because the answer should come out to be just x.

I multiplied everything by 8 to get it out of the denominator and simplified the numerator which left me with

cube root of {64x^3-432}
You multiplied everything by 8? Wouldn't it be easier just to cancel the "8" in the numerator with the "8" in the denominator?
 

What is a cube root?

A cube root is a mathematical operation that finds the number, when multiplied by itself three times, gives a given number. It is the inverse operation of cubing a number.

How do I solve a function with a cube root algebraically?

To solve a function with a cube root algebraically, you need to isolate the cube root term on one side of the equation and then raise both sides of the equation to the third power. This will eliminate the cube root and allow you to solve for the variable.

Can I use the quadratic formula to solve a function with a cube root?

No, the quadratic formula is only applicable to functions with quadratic terms. To solve a function with a cube root, you need to use the process of isolating the cube root and raising both sides to the third power.

What if the function has a cube root and other terms?

If the function has a cube root and other terms, you need to first isolate the cube root term and then use the appropriate algebraic operations to solve for the variable. Make sure to follow the order of operations and simplify as needed.

Are there any special rules or properties for solving functions with cube roots?

Yes, one important rule to remember is that the cube root of a negative number is a complex number. Also, when solving for the cube root of a variable, there could be more than one solution, so it is important to check your answer to make sure it satisfies the original function.

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