# Someone help me with this math question?

### 3 Answers

- PinkgreenLv 78 months ago
g^-1(x)=(5-kx)/3

=>

g(x)=(5-3x)/k

g(x^2)=2f(-x)

=>

[5-3(x^2)]/k=2(3x^2-5)

=>

5-3x^2=6kx^2-10k---(1)

(k=/=0)

=>

x^2=(10k+5)/(6k+3)

=>

x=+/-sqr[(10k+5)/(6k+3)]

=>

(10k+5)/(6k+3)>=0

=>

k>-1/2 or k<-1/2

[Note that k=-1/2=>

(10k+5)/(6k+3)=0/0 which

is uncertain; but (1) holds

=>k=-1/2 is acceptable]

=>

k can be any real number

but 0, or write k=R-{0}.

- Ian HLv 78 months ago
f(x) = 3x^2 – 5

2f(-x) = 2[3(-x)^2 – 5] = 6x^2 – 10 ........(1)

The next part assumes you meant inverse function rather than reciprocal.

The inverse of an inverse is the original function, so to find inverse of

Let g^-1(x) = y(x) = (5 – kx)/3

x = (5 – 3y)/k ....now swap roles

y = (5 – 3x)/k and interpret role-reversed as

g(x) = (5 – 3x)/k

g(x^2) = (5 – 3x^2)/k .........................(2)

For g(x^2) = 2f(-x) we have

(5 – 3x^2)/k = 6x^2 – 10

k = -(3x^2 - 5)/(6x^2 – 10) = -1/2

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- AlanLv 78 months ago
g^-1(x) = (5 -kx)/3

f(x) = 3x^2 - 5

g(x^2) = 2*f(-x)

so 2*f(-x) = 2* (3x^2 -5)

since (-x)^2 =x^2

2*f(-x) = 6x^2 -10

g^-1(x) = (5-kx)/3

so inverse of the inverse is

the original function

so if inverse is:

y = (5-kx)/3

swap x and y and solve for new y

x= (5-ky)/3

3x = 5-ky

ky= 5-3x

y = (5-3x)/k

g(x) = (5-3x)/k

g(x^2) = (5-3x^2)/k

so you want k such that

6x^2 -10 = (5-3x^2)/ k

6kx^2 -10k = 5- 3x^2

Equate like terms

6kx^2 = -3x^2

divide both sides by 3x^2

2k = -1

k = -1/2

checking

-10k = 5

divide both sides by -10

k = -5/10 = -1/2 (good ,

both answers must be the same)

so k = -1/2

checking the answer

g^(-1)(x) = (5+(x/2))/ 3

g^-1(x) = 5/3 + x/6

inverse of this

x= 5/3 +y/6

multiply by 6

6x = 10 + y

y = 6x-10

g(x) = 6x -10

g(x^2) = 6x^2 -10

f(x) =3x^2-5

2*f(-x) = 2*(3(-x)^2 -5) = 6x^2 -10

- ...Show all comments
Yes, it should say for all x.

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The question is ambiguous, it may be an equation of x & k

is a parameter. So it must be stated clearly that it is an identity in x.