Question

If f, g, h are real functions defined by $f\left(x\right)=\sqrt{x+1},g\left(x\right)=\frac{1}{x}$ and $h\left(x\right)=2x^2-3$, then find the values of (2f + g - h) (1) and (2f + g - h) (0).

Answer 1

We have
$f\left(x\right)=\sqrt{x+1},g\left(x\right)=\frac{1}{x}$
and $h\left(x\right)=2x^2-3$
Clearly, f(x) is defined for $x+1\ge 0$
$\Rightarrow x\ge -1\Rightarrow xϵ[-1,\infty ]$
g(x) is defined for $x\ne 0$
$\Rightarrow xϵR-\left\{0\right\}$
and h(x) is defined for all $xϵR$
$\therefore \text{Domain}\left(f\right)\cap \text{Domain}\left(g\right)\cap \text{Domain}\left(h\right)=[-1,\infty ]-\left\{0\right\}$
Clearly,
$2f+g-h:[-1,\infty ]-\left\{0\right\}\to R$ is given by
$(2f+g-h)\left(x\right)=2f\left(x\right)+g\left(x\right)-h\left(x\right)=2\sqrt{x+1}+\frac{1}{x}-2x^2+3\therefore (2f+g-h)\left(1\right)=2\sqrt{1+1}+\frac{1}{1}-2\times (1{)}^2+3=2\sqrt{2}+1-2+3=2\sqrt{2}+4-2=2\sqrt{2}+2$
and, (2f+g-h)(0) does not exist, it is not lies in the domain $xϵ[-1,\infty ]-\left\{0\right\}$.