Question

If f(×)= x^1/2-× and g(×) = 2x^3-×^1/2-×, find f(×)-g(×). Pleade ASAP

Answer 1

-2x^3 + 2x^(1/2)

Explanation:

step one- plug in the values of each in f(x) - g(x)

x^(1/2)-x - (2x^3-×^(1/2)-x) ; g(x) is in parentheses bc we need to distribute the negative

x^(1/2)-x-2x^3 + x^(1/2) + x ; combine like terms (letters with the same exponents) ; the x's cancel, and the x^(1/2) + x^(1/2) combine to form 2x^(1/2); the -2x^3 remains

you should get

-2x^3 + 2x^(1/2)

Answer 2

`f(x) - g(x) = 2(√(x)-x^3)`

Explanation:

We have the functions

`f(x)= x^{(1)/(2)}-x` and
`g(x) = 2x^3-x^(1)/(2)-x`

The operation
`f (x) -g (x)` is the subtraction of the function f(x) minus the function g(x). Thus

`f(x) - g(x) = x^{(1)/(2)}-x -(2x^3-x^(1)/(2)-x)`

Simplifying the expression we have left that:

`f(x) - g(x) = x^{(1)/(2)}-x -2x^3+x^(1)/(2)+x`

`f(x) - g(x) = 2x^{(1)/(2)}-2x^3`

`f(x) - g(x) = 2(√(x)-x^3)` Because
`x^{(a)/(n)} = \sqrt[n]{x^a}`