Question

F(x)= ax+3 ; `0≤x<1/2` f(x)= `sqrt((2x-1)/(x+15))` ; `1/2≤x≤1` f(x)= `(sqrt(x+3)-2)`/(x-1) ; `x>1` Find the value of a that makes f(x) continuous in `1/2` ?

Answer 1

`a=color(red)(-6)`

Explanation:

Since `f(x)=sqrt((2x-1)/(x+15))` for `1/2 <= x < 1`

at `x=1/2`
`color(white)("XXX")f(1/2)=sqrt((2 * 1/2 -1)/(1/2+15))=color(magenta)0`

If `f(x)=color(blue)ax+3` for `x < 1/2`

to be continuous
as `x rarr 1/2` then `color(blue)ax+3 rarr color(magenta)0`

that is, as `xrarr 1/2`
`color(white)("XXX")color(blue)a * 1/2 +3 = color(magenta)0`

`color(white)("XXX")color(blue)a+6 =color(magenta)0`

`color(white)("XXX")color(blue)a=color(red)(-6)`