Question

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

Answer 1

Answer:  

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

Step-by-step 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

Answer:

$f(x) - g(x) = 2(\sqrt{x}-x^3)$

Step-by-step explanation:

We have the functions

$f(x)= x^{\frac{1}{2}}-x$  and   $g(x) = 2x^3-x^\frac{1}{2}-x$

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

$f(x) - g(x) = x^{\frac{1}{2}}-x -(2x^3-x^\frac{1}{2}-x)$

Simplifying the expression we have left that:

$f(x) - g(x) = x^{\frac{1}{2}}-x -2x^3+x^\frac{1}{2}+x$

$f(x) - g(x) = 2x^{\frac{1}{2}}-2x^3$

$f(x) - g(x) = 2(\sqrt{x}-x^3)$  Because $x^{\frac{a}{n}} = \sqrt[n]{x^a}$