Find f^-1(x)....

[samistumbo]

Need help.... don't know where to start.
Question:
Let f(x)= √(16- x²), 0 ≤ x ≤ 4.

(a) Find f^-1.

f^-1(x)= ___________ (for 0 ≤ x ≤ 4)

How is it related to f?

 

[lookagain]

Need help.... don't know where to start.
Question:
Let f(x)= √(16- x²), 0 ≤ x ≤ 4.

(a) Find f^-1.

f^-1(x)= ___________ (for 0 ≤ x ≤ 4)

How is it related to f?

An inverse function is the original function reflected across the line y = x,
and vice versa.

Sketch the given function.

If, when the graph is reflected across the line y = x, and you get the same
graph, then the inverse function is equal to the given function.


Note: The domain of the inverse function (interval notation-wise),
is the same as the range of the given function (interval notation-wise),
and vice versa.

 

[burakaltr]

Need help.... don't know where to start.
Question:
Let f(x)= √(16- x²), 0 ≤ x ≤ 4.

(a) Find f^-1.

f^-1(x)= ___________ (for 0 ≤ x ≤ 4)

How is it related to f?

y=sqrt(16-x**2)
y**2=16-x**2
16-y**2=x**2
x=sqrt.(16-y**2)

f-1(x)=sqrt(16-x**2)

 

[lookagain]

y=sqrt(16-x**2) $\displaystyle . . . . . . semicircle \ in \ Quadrants \ I \ and \ II$

y**2=16-x**2

16-y**2=x**2

x=sqrt.(16-y**2) $\displaystyle . . . . . . semicircle \ in \ Quadrants \ I \ and \ IV$

f-1(x)=sqrt(16-x**2)

$\displaystyle y^2 = 16 - x^2 \ is \ a \ whole \ circle, \ but \ the \ given$

$\displaystyle function \ is \ a \ \text{quarter-circle} \ in \ the \ first \ quadrant.$

$\displaystyle y \ = \sqrt{16 - x^2} \ is \ a \ semicircle \ in \ the \ first \ and \ second \ quadrants.$


Regarding these three, the whole circle is a relation that is not a function. The semicircle
is a function, but it is not a one-to-one function. The quarter-circle is a function, and it is
a one-to-one function.


$\displaystyle With \ the \ given \ restriction \ on \ the \ semicircle, \ \ 0 \le x \le 4,$

$\displaystyle that \ makes \ it \ a \ \text{quarter-circle}.$


burakltr, you are missing the restriction on x.

 

[burakaltr]

$\displaystyle y^2 = 16 - x^2 \ is \ a \ whole \ circle, \ but \ the \ given$

$\displaystyle function \ is \ a \ \text{quarter-circle} \ in \ the \ first \ quadrant.$

$\displaystyle y \ = \sqrt{16 - x^2} \ is \ a \ semicircle \ in \ the \ first \ and \ second \ quadrants.$


Regarding these three, the whole circle is a relation that is not a function. The semicircle
is a function, but it is not a one-to-one function. The quarter-circle is a function, and it is
a one-to-one function.


$\displaystyle With \ the \ given \ restriction \ on \ the \ semicircle, \ \ 0 \le x \le 4,$

$\displaystyle that \ makes \ it \ a \ \text{quarter-circle}.$


burakltr, you are missing the restriction on x.

Yes Sir, thank you