Can anyone help me find this inverse?

Question

f(x)=square root(2x+5)

i got y=2x-5/2 but i dont know if i did it right thanks

Answer 1

that is absolutely wrong
y=sqrt(2x+5)
square both sides
y^2=2x+5
y^2-5=2x
[y^2-5]/2=x
you want to find the inverse you justswitch x and y
inverse is y={x^2-5]/2 that is the answer
let check. i x=2 we put in the original function we have sqrt(2*2+5)=sqrt(9)=3.

if the inverse function is correct so f^-1(3)=2
let put x=3 into the inverse function
y=(3^2-5)/2= [9-5]/2= 4/2= 2. yeah that is correct function you may check your answer again somehow you messed up. good luck

Answer 2

Just swap x and f(x) and solve for f(x):
f(x)=sqrt(2x+5)
x = sqrt(2*f inv (x) + 5)
x^2 = 2*f inv (x) + 5
(1/2)x^2 = f inv (x) + 5
(1/2)x^2 - 5 = f inv (x).

And you're done! Except that the sqrt function has restricted domain, so 2x+5 >= 0; x >= -5/2 in f(x). Thus, in f inverse, x >= sqrt(5). (Most texts write f followed by a superscript of -1, but that's hard to do here.)

Answer 3

y = sqr(2x+5)

exchange x and y
x = sqr(2y+5)

x^2 = 2y + 5

x^2 - 5 = 2y

y = 1/2x^2 -5/2

the original equation has only onve curve but the inverse has two curves, we have to find the domain.

since the the original moves up, the domain for the inverse is
x >/ 0

ans: 1/2x^2 -5/2 , x >/ 0

Answer 4

You're just missing the 1/2x^2...just note that depending on the strictness of your teacher, an absolute value may be required as the range of sqrt is x >= 0. so f^-1(x) = 1/2|x|^2 - 5/2

To solve, in n the original equation replace f(x) with x and replace x with f^-1(x). Then just isolate f^-1(x).

To check, plug in f^-1(x) into f(x) and make sure you get x.

Answer 5

y=(x^2)/2-5/2



x= Sqrt[2y+5]
x^2=2y + 5
x^2-5=2y
(x^2)/2-5/2=y