For the function f:[1,) [1,) defined by f(x)=(x-1)^4+1, among the two statements: (I) The …

Question

For the function $f:[1,\infty)	o [1,\infty)$ defined by $f(x)=(x-1{)}^{4}+1$, among the two statements:

(I) The set $S={x\in [1,\infty):f(x)={f}^{-1}(x)}$ contains exactly two elements, and

(II) The set $S={x\in [1,\infty):f(x)={f}^{-1}(x+1)}$ is an empty set,

Accepted Answer

For

$f:[1,\infty)	o [1,\infty),\, f(x)=(x-1{)}^{4}+1,$

we first find the inverse function.

Since for $x\ge 1$, the function $(x-1{)}^{4}+1$ is strictly increasing, $f$ is one-one and onto from $[1,\infty)$ to $[1,\infty)$.

Let

$y=(x-1{)}^{4}+1.$

Then

$y-1=(x-1{)}^{4}.$

Since $x\ge 1$, we take the positive fourth root:

$x-1=(y-1{)}^{1/4}.$

Hence

${f}^{-1}(y)=1+(y-1{)}^{1/4}.$

So

${f}^{-1}(x)=1+(x-1{)}^{1/4}.$

Now check the two statements one by one.

Statement (I)
We need to solve

$f(x)={f}^{-1}(x).$

That is,

$(x-1{)}^{4}+1=1+(x-1{)}^{1/4}.$

So

$(x-1{)}^{4}=(x-1{)}^{1/4}.$

Put

$t=x-1\, t\ge 0.$

Then the equation becomes

${t}^{4}={t}^{1/4}.$

Now solve:

${t}^{4}-{t}^{1/4}=0.$

This gives

${t}^{1/4}({t}^{15/4}-1)=0.$

Hence

$t=0\, \, {t}^{15/4}=1.$

Since $t\ge 0$, from ${t}^{15/4}=1$ we get

$t=1.$

Therefore

$x-1=0\Rightarrow x=1,$

or

$x-1=1\Rightarrow x=2.$

So the set

$S={x\in [1,\infty):f(x)={f}^{-1}(x)}$

contains exactly two elements:

$S={1,2}.$

Thus, Statement (I) is TRUE.

Statement (II)
We need to solve

$f(x)={f}^{-1}(x+1).$

Now

$f(x)=(x-1{)}^{4}+1,$

and

${f}^{-1}(x+1)=1+((x+1)-1{)}^{1/4}=1+{x}^{1/4}.$

So the equation becomes

$(x-1{)}^{4}+1=1+{x}^{1/4}.$

Hence

$(x-1{)}^{4}={x}^{1/4}.$

Again put

$t={x}^{1/4},\, t\ge 1x\ge 1.$

Then

$x={t}^{4}.$

So the equation becomes

$({t}^{4}-1{)}^{4}=t.$

Instead of solving this directly, test simple values.

For $x=1$:

$f(1)=(1-1{)}^{4}+1=1,$

and

${f}^{-1}(1+1)={f}^{-1}(2)=1+(2-1{)}^{1/4}=2.$

So $x=1$ is not a solution.

For $x=2$:

$f(2)=(2-1{)}^{4}+1=2,$

and

${f}^{-1}(2+1)={f}^{-1}(3)=1+(3-1{)}^{1/4}=1+{2}^{1/4}
e 2.$

So $x=2$ is not a solution.

Now define

$g(x)=(x-1{)}^{4}-{x}^{1/4},\, x\ge 1.$

We need to check whether $g(x)=0$ has any solution.

Observe:

$g(1)=0-1=-1and

$g(2)=1-{2}\^{1/4}while

$g(3)=16-{3}^{1/4}>0.$

Since $g$ is continuous, there is at least one root in $(2,3)$.

So the set

$S={x\in [1,\infty):f(x)={f}^{-1}(x+1)}$

is not empty.

Thus, Statement (II) is FALSE.

Therefore, the correct option is

Solutions · 1