Answer
a) $f(0)=0$
(b) $f(1)=\sqrt {2}$
(c) $f(-1)=0$
(d) $f(-x)= \sqrt {x^2-x}$
(e) $-f(x)=-\sqrt {x^2+x}$
(f) $f(x+1)=\sqrt {x^2+3x+2}$
(g) $f(2x)=\sqrt {4x^2+2x}$
(h) $f(x+h)=\sqrt {x^2+2xh+h^2+x+h}$
Work Step by Step
Given $f(x)=\sqrt {x^2+x}$
a) To Evaluate $f(0)$ in the given function, substitute $0$ in the place of $x$.
$f(0)=\sqrt {0^2+0}=\sqrt {0^2}=0$
(b). To Evaluate $f(1)$ in the given function, substitute $1$ in the place of $x$.
$f(1)=\sqrt {1^2+1}=\sqrt {1+1}=\sqrt {2}$
(c). To Evaluate $f(−1)$ in the given function, substitute $−1$ in the place of $x$.
$f(-1)=\sqrt {(-1)^2-1}=\sqrt {1-1}=\sqrt {0}=0$
(d) To Evaluate $f(−x)$ in the given function, substitute $−x$ in the place of $x$.
$f(-x)= \sqrt {(-x)^2+(-x)}=\sqrt {x^2-x}$
(e) To Evaluate $−f(x)$ in the given function,
$-f(x)=-\sqrt {(x)^2+x}=-\sqrt {x^2+x}$
(f) To Evaluate $f(x+1)$ in the given function, substitute $x+1$ in the place of $x$.
$f(x+1)=\sqrt {(x+1)^2+(x+1)}=\sqrt {x^2+2x+1+x+1}=\sqrt {x^2+3x+2}$
(g) To Evaluate $f(2x)$ in the given function, substitute $2x$ in the place of $x$.
$f(2x)=\sqrt {(2x)^2+2x}=\sqrt {4x^2+2x}$
(h) To Evaluate $f(x+h)$ in the given function, substitute $x+h$ in the place of $x$.
$f(x+h)=\sqrt {(x+h)^2+(x+h)}=\sqrt {x^2+2xh+h^2+x+h}$
