Question

Which function does NOT have an inverse?
A)f(x) = 2x + 10, x ≥ -5
B)f(x) = 2x + 10, x ≥ 0
C)f(x) = 2x2 + 10, x ≥ -5
D)f(x) = 2x2 + 10, x ≥ 0

Answer 1

Okay, first things first, linear functions always have inverses, so scratch A and B.

Answer 2

okay
the last two are squared btw they didnt show up

Answer 3

Yeah... they both have the same function, but different domains... that's gotta be tough ^.^

Answer 4

i needed an explination

Answer 5

not sarcasm.

Answer 6

f(x) = 2x^2 + 10
\[\large y = 2x^2 + 10\]So what I want you to do is switch x and y
\[\large x = 2y^2 + 10\]
And make y alone.

Answer 7

isnt it \[\sqrt{5-x}=y\]

Answer 8

No.

Answer 9

first i subtracted 10

Answer 10

10-x=2y\[10-x=2y^{2}\]

Answer 11

2nd one

Answer 12

then i divided both sides by 2

Answer 13

\[10-x \div 2 =y ^{2}\]

Answer 14

then i square rooted both sides

Answer 15

for a function to have an inverse it must be one-one and onto type. I trust you know the two requirements?

Answer 16

yes

Answer 17

So, put your set of requirements, the correct out to be D because of onto type of function.

Answer 18

I mean, D ought to be invertible. *pardon*

Answer 19

@PeterPan Inverting a function \(f(x)\), you find \(f^{-1}(x)\)

Answer 20

C is not of onto type, and thus it is a non-invertible function. (I may be wrong here, relations and functions is not one of my better topics)

Answer 21

@natalia517 you are on the right track. follow @NeetziD

Answer 22

yeah, and that still involves switching x and y, and then isolating y. The resulting function of x would be f^-1(x)

Answer 23

alright thank you for helping me again neetziD

Answer 24

Don't mention. ;)

Answer 25

check:
for option C) \[f(x)=y=2x^2+10,\qquad\qquad\forall\quad x\ge-5\text{and}\;0\le y<\infty\\
\frac{y-10}{2}=x^2\implies x=\pm\sqrt{y-10\over2}
\]
this limits the possible range for "y"...

Answer 26

but you will notice that the domain of \(f(x)\) is \(10\le y<\infty\)

Answer 27

i meant, the "range"

Answer 28

and for the inverse, it is no longer one-to-one correcpondence and the difinition of function gets mangled. so cannot be called a function anymore

for the D) option, though it looks just the same, "x"can take only the POSITIVE values and maintains the one-to-one correspondence. and its inverse is a valid function.