1. Determine whether or not the given function is one-to-one and, if so, find th

Question

1. Determine whether or not the given function is one-to-one and, if so, find the inverse. If f(x)=(4x+7)^4 has an inverse, give the domain of f^1.

a) f^1(x)=(7x^(1/4))/4; domain: (0,)

b) f^1(x)=(7+4x)^(1/4); domain: (,7/4)

c) Not one-to-one

d) f^1(x)=(x^(1/4)7)/4; domain: (,)

e) f^1(x)=(7x^(1/4))/4; domain: (7/4,)

2. Determine whether or not the given function is one-to-one and, if so, find the inverse. If f(x)=(6+3x^2)^7 has an inverse, give the domain of f^1.

a) f^1(x)=((1/3)x^(1/7)2)^(1/2); domain: (,)

b) f^1(x)=(6+3x^2)^(1/7); domain: (0,)

c) Not one-to-one

d) f^1(x)=(6+3x^2)^(1/7); domain: (,)

e) f^1(x)=((1/3)x^(1/7)2)^(1/2); domain: (0,)

Explanation / Answer

for a function should be one to one

for 2 different values of x the values of y should be also different.

(1)

given function

f(x)=(-4x+7)4

so, for checking of this

we have to check

f(x)=f(y) ===> x=y

so,

f(y)=(-4y+7)4

so

(-4x+7)4=(-4y+7)4

(-4x+7)=((-4y+7)4)1/4   (since even power)

so, from here we can see that x can take 2 values where y is same

so,

this function is not "one -to- one"

so, no inverse exist.

option (c)

(b)

given function is

f(x)=(6+3x2)7

similarly

f(y)=(6+3y2)7

so,

(6+3x2)7=(6+3y2)7

since it has odd power so, it will get cancelled

(6+3x2)=(6+3y2)

3x2=3y2

x2=y2

so

x=y or x=-y

so, here too it has 2 values

so, this function is also not "one to one"

so no inverse exist.

option(c)

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