Question

If `f( x ) = \sqrt { 2x + 16}`, what is `f(0), f(2), f(4), f(-6)`?

Answer 1

`f(0)=+-4`, `f(2)=+-2sqrt5`, `f(4)=+-2sqrt6` and `f(-6)=+-2`

Explanation:

When you have to find `f(0), f(2),f(4)" or f(-6)` for a given `f(x)`,

what you do is substitute `x` for `0,2,4" or "-6` in `f(x)` and get corresponding value of `f(x)` by carrying out respective arithmetic operations.

Here we have `f(x)=sqrt(2x+16)`, but before we find vale of `f(x)`, note that as we have square root function, we can have both negative as well as positive values. Further, if `2x+16<0`, function is not defined.

hence `f(0)=sqrt(2xx0+16)=sqrt16=+-4`

`f(2)=sqrt(2xx2+16)=sqrt(4+16)=sqrt20=sqrt(2xx2xx5)=+-2sqrt5`

`f(4)=sqrt(2xx4+16)=sqrt(8+16)=sqrt24=sqrt(2xx2xx6)=+-2sqrt6`

and `f(-6)=sqrt(2xx(-6)+16)=sqrt(-12+16)=sqrt4=+-2`