Question

If `f(x)=5x`, `g(x)=1/x`, and `h(x)=x^3`, how do you find `(5h(1)-2g(3))/(3f(2))`?

Answer 1

`13/90`, or `0.1bar4`

Explanation:

`(5h(1) - 2g(3))/(3f(2))`

Now let's plug in the values for those functions:

`(5*(1^3) - 2*(1/3))/(3*(5*2))`

`1^3 = 1`, and since any number times `1` is the same, we can remove it.

`(5 - 2*(1/3))/(3*(5*2))`

Now we can simplify `2*(1/3)` to `2/3`

`(5 - 2/3)/(3*(5*2))`

And we can multiply the numbers in the denominator `3*5 = 15`, and `15 *2 = 30`

`(5 - 2/3)/30`

Now let's try and get rid of that ugly fraction in a fraction! We can do that by multiplying the entire fraction by `3/3` which is also equal to `1`, and like we said above, we're not actually changing the value by doing that!

`(5 - 2/3)/30 * 3/3`

Multiplying through on both sides:

`(5*3 - (2/3)*3)/(30*3)`

Now simplify the multiplications:

`(15 - 2)/90`

Now subtract in the numerator... and...

`13/90`

There you have it!

Answer 2

`13/90`

Explanation:

`(5h(1)-2g(3))/(3f(2))`
`=(5(1)^3-2(1/3))/(3(5*2)`
`=(5-2/3)/(3*10)`
`=(13/3)/30`
`=13/3*1/30`
`=13/90`