# plz try this simple quest..the ans is different as i calculated.

• Jan 8th 2010, 10:06 PM
redcherry
plz try this simple quest..the ans is different as i calculated.
find natural domain and range for each function:
a) G(x)=(x^2-2*x+5)^(1/2)

b)h(x)=1/(1-sin(x))

c)H(x)=(sin(x^(1/2)))^(-2)
• Jan 8th 2010, 10:15 PM
mr fantastic
Quote:

Originally Posted by redcherry
find natural domain and range for each function:
a) G(x)=(x^2-2*x+5)^(1/2)

b)h(x)=1/(1-sin(x))

c)H(x)=(sin(x^(1/2)))^(-2)

• Jan 8th 2010, 10:29 PM
redcherry
i just use this formula to find the roots which men the values of x..
(-b+(b^2-4*a*c))/2*a and got the ans x=1+4i and x=1-4i
but the ans for the domain is x=real number and the range y>=2.
• Jan 8th 2010, 10:36 PM
mr fantastic
Quote:

Originally Posted by redcherry
i just use this formula to find the roots which men the values of x..
(-b+(b^2-4*a*c))/2*a and got the ans x=1+4i and x=1-4i
but the ans for the domain is x=real number and the range y>=2.

What you posted is not what I consider detailed working.

a) Domain: Solve $\displaystyle x^2-2x+5 \geq 0$.

b) Domain: All real numbers except those satisfying 1-sin(x) = 0.

c) Domain: All positive real numbers except those satisfying $\displaystyle \sin(\sqrt{x}) = 0$.

The details are left for you. I suggest you get ranges by drawing graphs. For a) note that $\displaystyle y=(x^2-2x+5)^{1/2} \Rightarrow y^2 = x^2 - 2x + 5$ and you should recognise a branch of a hyperbola.
• Jan 8th 2010, 11:09 PM
redcherry
ok..
for b) i got the domain x is not equal (2*n+1/2)*Pi but i didnt got the range
for c) the domain x is not equal (n*Pi)^2 and also didnt get the range..
• Jan 9th 2010, 02:37 AM
mr fantastic
Quote:

Originally Posted by redcherry
for b) i got the domain x is not equal (2*n+1/2)*Pi but i didnt got the range
for c) the domain x is not equal (n*Pi)^2 and also didnt get the range..

Draw the graphs. eg. b) Draw the graph of $\displaystyle f(x) = -\sin (x) + 1$ and then use the standard properties of reciprocal functions to draw $\displaystyle h(x) = \frac{1}{f(x)}$.
• Jan 9th 2010, 02:39 AM
Prove It
Quote:

Originally Posted by mr fantastic
Draw the graphs. eg. b) Draw the graph of $\displaystyle f(x) = -\sin (x) + 1$ and then use the standard properties of reciprocal functions to draw $\displaystyle h(x) = \frac{1}{f(x)}$.

Alternatively, to find the natural range of each function, find the domain of each inverse function...