Question

please let f(x)= x [ 1/2 (1-(X^2))^1/2 (-2x) ] + sqroot (1-(x^2))
a. For what values is f(x)=0?
b. For what values is f(x) undefined?
c. For what values is f(x)<0?
d. For what values is f(x)>0?

Answer 1

\[f(x) = x[\frac{1}{2}(1-x^2)^{\frac{1}{2}}(-2x)]+\sqrt{1-x^2}\]
is this the equation?

Answer 2

yes it is

Answer 3

\[\text{A. }\ x = 1\ \ or\ \ x = -1\]

Answer 4

\[x \left[ \frac{1}{2} (1-x^2)^\frac{1}{2} (-2x) \right] + \sqrt{1-x^2} \]\[ = x \left[ - x\sqrt{(1-x^2)} \right] + \sqrt{1-x^2} \]\[ = -x^2\sqrt{1-x^2} + \sqrt{1 - x^2}\]\[ = (1 - x^2)\sqrt{1 - x^2}\]\[ = (1-x^2)^{\frac{3}{2}}\]

Part A:
\[(1-x^2)^{\frac{3}{2}} = 0\]\[x^2 = 1\]\[x = \pm1\]
Part B:
f(x) is undefined when:
\[(1 - x^2) < 0\]\[x^2 > 1\]\[x < -1 \space\vee\space x > 1\]

Answer 5

\[\text{B. }\ x >1\ \ or\ \ x < 1\]
\[\text{C. }\phi\]
\[\text{D. } x < 1\ or\ x > -1\]

Answer 6

in other words for D. -1 < x < 1

Answer 7

Part C:
I think no values of x satisfy this, as the square root will only give positive values:
Part D:
Any values between -1 and 1 exclusive. Because 1 - x^2 must be greater than 0.

Answer 8

thanks a lot . appreciate it