calculate the exact value of x when f(x)=6

Question

suppose f(x)=6/5-x^3

calculate the exact value of x when f(x)=6

 

so: 6=6/5-x^3

6=6(5-x^3)

6=30-6x^3

-6x^3=6-30

-6x^3=-24

-6x^3+24=0

6x^3-24=0

6(x^3-4)=0

x^3-4=0

x^3=4

x=cube root 4

x=1.5874

needs to be a fraction form

but x=4/3 is incorrect

where did I go wrong?

 

Answer 1

Virginia,

 

Based on how you solved this problem you ended up with:

 

x=³√4

 

I believe what they are asking by fractional form is that what you have is:

 

4^1/3 or if they want this one step further you could have 2^2/3.

 

I hope this helps. I think it may be what they mean by "fractional form"

 

-Adam

Answer 2

Based on what your steps are, I think it is f(x) = 6/(5 -x³)

If f(x) = 6, substitute:

 

If 6 = 6/(5 - x³), then make it simpler.  5 - x³ must equal 1 for the fraction 6/(5 - x³) to equal 6:

 

5 - x³ = 1

-x³ = -4

-x = -³√4

x = ³√4

 

This is the exact value, since any step further would make it irrational, and therefore not exact.