Inverse trig (1 Viewer)

Aysce · May 17, 2012

Question:

Find the domain of y=arcsin(x^2)

barbernator · May 17, 2012

-1 < x^2 < 1

-1 < x < 1

all inequalities are equal tos

Alkenes · May 17, 2012

barbernator said:
-1 < x^2 < 1

-1 < x < 1

shouldn't that be greater and EQUAL and less than and EQUAL

Aysce · May 17, 2012

barbernator said:
-1 < x^2 < 1

-1 < x < 1

all inequalities are equal tos

I don't really understand how you jumped to the second step?

Alkenes · May 17, 2012

u square root right and left hand side of the inequality to get x in the middle

Aysce said:
I don't really understand how you jumped to the second step?

barbernator · May 17, 2012

Aysce said:
I don't really understand how you jumped to the second step?

well if x > 1 or x< -, x^2 > 1.
x= + -1, x^2=1
x < 1 or x > -1, -1 < x^2 < 1 but we know x^2 is positive so 0 < x^2 < 1 which fits our bounds.

again, some inequality signs are equal to.

Aysce · May 17, 2012

Alkenes said:
u square root right and left hand side of the inequality to get x in the middle

Yeah but in 3u you can't square negative numbers.

idgi

Alkenes · May 17, 2012

yeah thats what i was confused about this question.... the left hand side would = i^(1/2)

Aysce said:
Yeah but in 3u you can't square negative numbers.

idgi

barbernator · May 17, 2012

Aysce said:
Yeah but in 3u you can't square negative numbers.

idgi

regardless. we know that x^2 is positive, so it is greater than 0. when you take the square root of x^2 < 1. you get a positive and a negative solution. x < 1, x > -1.

barbernator · May 17, 2012

Alkenes said:
yeah thats what i was confused about this question.... the left hand side would = i^(1/2)

LHS would be just i btw bro

Alkenes · May 17, 2012

yeah square root of i = i

barbernator said:
LHS would be just i btw bro

Aysce · May 17, 2012

barbernator said:
regardless. we know that x^2 is positive, so it is greater than 0. when you take the square root of x^2 < 1. you get a positive and a negative solution. x < 1, x > -1.

Ah okay thanks. I might have some more questions later but i'll try them first.

Timske · May 17, 2012

you don't use algebra to solve these

barbernator · May 17, 2012

Alkenes said:
yeah square root of i = i

dude, get your facts straight. if we are taking the square root of -1, we get sqrt-1=i

Sanjeet · May 17, 2012

Let's try a different approach
$$The domain of $y=sin^{-1}(x^2)$ is equal to the range of its inverse function, so let's find the inverse$ \\ x=sin^{-1}(y^2)\\ sinx=y^2\\y=\sqrt{sinx} \\ $We know the values of $sinx$ range from $ -1\le x \le1,$ however as it is in a square root y must be $\ge0.$ Therefore the range of this function is $0\le y\le1,$ and thus the domain of its inverse is $0\le x \le1$

barbernator · May 17, 2012

Sanjeet said:
Let's try a different approach
$$The domain of $y=sin^{-1}(x^2)$ is equal to the range of its inverse function, so let's find the inverse$ \\ x=sin^{-1}(y^2)\\ sinx=y^2\\y=\sqrt{sinx} \\ $We know the values of $sinx$ range from $ -1\le x \le1,$ however as it is in a square root y must be $\ge0.$ Therefore the range of this function is $0\le y\le1,$ and thus the domain of its inverse is $0\le x \le1$

the domain is -1 < x < 1 . you did not take both positive and negative solutions to the quadratic

again, inequalities are also equal to

Sanjeet · May 17, 2012

my bad!!

SpiralFlex · May 17, 2012

$Graphically we can convey the meaning of arcsin(x^2)$

$For an hypothetical curve, y=f(x). Let us consider how we would represent y=f(x^2)$

$$If we substitute a value numerically equal, but opposite in sign, we would get the same output. For example, when x=-4, ie. f(16), x=4, ie. f(16). There would be symmetry here about the y axis.$

$We can also consider the x values to be going rapidly.$

$$The only real point that would remain consistent is f(0) and f(0 squared). Also, f(1) and f(1 squared). For both graphs of y=f(x) and y=f(x squared). From here you can use a mental plot of the shape of the graph, which is a rice bowl shape.$

[HR][/HR]

$But all we really need is to be able to do this algebraiicly, which is what barbie, I mean barbenator has shown.$

Sanjeet · May 17, 2012

lmao^

SpiralFlex · May 17, 2012

Sanjeet said:
lmao^

Why you laugh at Spiral?

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