Question

Need help: If h(x) = x + x^(1/2), then find h^(-1)(6).

Answer 1

the answer you want to find is 6/h??

Answer 2

The book just says h to the power of -1 and then times (6)

Answer 3

Im looking at an example in the book where they did something like this, and they took the derivative of the whole equation, and then somehow found out what the inverse H of a # is.

Answer 4

do you know how to find the inverse of x + x^(1/2)

Answer 5

ah./... it means invers funtion..

Answer 6

Replace the X with y's and solve for Y

Answer 7

The question is under the inverse function section of the math book. I think I solved it. Here I will write out my answer and you guys tell me if it's right.

Answer 8

it has to do something with inverse function i think

Answer 9

According to their example, they took the derivative of their equation. So I did the same thing with this. H(x) = 1 + 1/2x^(-1/2). Graphed the original equation, swapped the points around to get the inverse function, and got what the inverse of h(6) is. I got 4.

So, using a theorem, I put that whole equation under a 1 on the nominator. Replaced the X with the 4 and solved it. I got 1 as my final answer

Answer 10

Here is my work: = 1 + (1/2) x ^ (-1/2).
= 1/(1+(1/2)x^(-1/2))
= 1/2x^(1/2) / 1
= 1/2x^(1/2)
inverse of h(6) is 4.
1/2(4)(1/2) = 1

Answer 11

\[h(x)=x+\sqrt{x}\]

Answer 12

they want u to evaluate\[h^{-1}(6)\]

Answer 13

The question is not clear, there are not any other instructions.

Answer 14

h^(-1) is the inverse function of h

just a neat point

if (x,y) lies on the the h then (y,x) lies on the h^(-1)

Answer 15

h(4)=6 so (4,6) lies on the h so (6,4) lies on h^(-1) and we can write\[h^{-1}(6)=4\]

Answer 16

yeah, that's what I got when I graphed it.

Answer 17

or u can replace x and y and evaluate the inverse function itself

Answer 18

Ah I see, makes sense

Answer 19

\[x=y+\sqrt{y}\]\[x=y+\sqrt{y}+\frac{1}{4}-\frac{1}{4}\]\[x=(\sqrt{y}+\frac{1}{2})^2-\frac{1}{4}\]\[x+\frac{1}{4}=(\sqrt{y}+\frac{1}{2})^2\]i think u got it from here

Answer 20

I don't quite get what you did there. How did you get the 1/4 and the -1/4?

Answer 21

thats nothing actually\[\frac{1}{4}-\frac{1}{4}=0\]i made it from nothing

Answer 22

Hmm still lost, sorry. The how did you get the 1/2, and the 1/4? I mean even after the second step

Answer 23

Then how*