I need help finding the inverse of a function?

Question

-2 sqr root (3+x) -4

-2 (sqrt of 3+x) -4

Answer 1

f(x) = -2sqrt(3 + x) - 4

Before we find the inverse of this function, we absolutely require the range of this function. The reason why is because the range of this function will be the domain of the inverse.

Note that sqrt(3 + x) >= 0. Multiplying both sides by -2,

-2sqrt(3 + x) <= 0

Subtracting both sides by -4,

-2sqrt(3 + x) - 4 <= -4

Therefore, the range of this function is (-infinity, -4]

Back to the inverse; if
f(x) = -2sqrt(3 + x) - 4

Let f(x) = y. Then

y = -2sqrt(3 + x) - 4

Now, swap the x and y variables, and then solve for y.

x = -2sqrt(3 + y) - 4
x + 4 = -2sqrt(3 + y)

Square both sides,

(x + 4)^2 = 4(3 + y)

Expand the right hand side,

(x + 4)^2 = 12 + 4y

And now, solve for y.

y = (1/4)[(x + 4)^2 - 12], or

y = (1/4)(x + 4)^2 - 3

At this point, we make y = f^(-1)(x), and

f^(-1)(x) = (1/4)(x + 4)^2 - 3

So this is the inverse _WITH ONE LITTLE CAVEAT_ ... the domain of this function is the range of f(x); in order for this to be the true inverse of the original function f(x), we have to force the domain of this to be (-infinity, -4]. So the complete answer is:

f^(-1)(x) = (1/4)(x + 4)^2 - 3 , for real numbers x <= -4

Answer 2

y = -2 sqr root (3+x) -4
y + 4 = -2 sqr root (3+x)
(y + 4)/2 = - sqr root (3+x)
(y + 4)^2 / 4 = 3 + x
x = -3 + (y + 4)^2 / 4
Interchange x , y:
The inverse function is
y = -3 + (x + 4)^2 / 4
Now you have to find the domain of this function:
Observe that the given function y < -4. So the domain of the inverse is x < -4.
Thus the inverse function is

y = -3 + (x + 4)^2 / 4, where x < -4

Answer 3

Write it as

y = -2 • √(3+x) - 4, and then interchange x and y

x = -2 √(3+y) - 4, and solve for y:

(x + 4) = -2 √(3+y)
(x + 4)² = 4(3 + y)
(x+4)²/4 = 3 + y
y = (x+4)²/4 - 3.

f'(x) = (x+4)²/4 - 3.

But don't forget about domain and range. The original function's domain was [-3, ∞) and its range was (-∞, -4]. So the domain of the inverse is (take a good look at the function) all reals (restricted to (-∞, -4]) and its range is [-3, ∞).

Answer 4

You've got this function:

y = -2 root(3+x) - 4

There are two steps:
1) Switch x and y
2) Solve for y

x = -2 root(3+y) - 4

If that looks intimidating, you're really doing PEMDAS (order of operations) in reverse order. Another way to think about it is unpealing an onion - in what order should you peel off the onions to get y?

x + 4 = -2 root(3 + y)
(x+4)/-2 = root(3 + y)
((x+4)/-2)^2 = 3 + y
((x+4)/-2)^2 - 3 = y <---- inverse

If you look carefully, you'll see that the operations from before are all reversed from the original equation, and in the reverse order (instead of subtracting 4 from x last, 4 is added to x first).

Answer 5

y = -2 sqr root (3+x) - 4
y+4 = -2 sqr root (3+x)
(y+4) / -2 = sqr root (3+x)
--------------------------------- to the power of 2
(y+4)^2 / 4 = (3+x)
{(y+4)^2 / 4} - 3 = x

so the inverse of that function is :
y = {(x+4)^2 / 4} - 3

Answer 6

to find the inverseof a function write the equation in y = f(x) form. so y = -2 sqr root (3+x) -4.
then solve switch x and y.
y = -2 sqr root (3+y) -4.
solve for y plug the new value for y back into the orriginal equation.

Answer 7

I think this works....

y = -2*sqrt(3*x)-4
y+4 = -2*sqrt(3*x)
(y+4)/-2 = sqrt(3*x)
[(y+4)/-2]^2 = 3*x
(1/3)*[(y+4)/-2]^2 = x

And square roots are 1-to-1, so the inverse exists. Hope this helps.

Answer 8

y=-2 sqr root (3+x)-4

Switch x and y.
Solve for y.
That's the inverse. =)

Answer 9

1) Assume f = -2 * sqrt( 3 + x ) - 4 and the problem is to find x = g(f).

2) sqrt(x+3) = -(f+4)/2;

3) x+3 = (f+4)^2 / 4

4) x = ( f^2 + 16 + 8f ) / 4 - 3

5) x = ( f^2 + 8f + 4) / 4;

Answer 10

clarify your question please with parentheses.