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Mathematics 13 Online
OpenStudy (anonymous):

how would i figure out if f(x) = x4 + 2; 0 ≤ x ≤ 6 is a "one to one" function?

OpenStudy (anonymous):

you could find the inverse, if the inverse if a function over the domain. then f(x) is one-to-one

OpenStudy (anonymous):

the domain being 0 ≤ x ≤ 6 ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

if f(x) outputs only 1 value for each x in the domain, then f(x) is one-to-one

OpenStudy (anonymous):

normally f(x)=x^4+2 would not be one-to-one, because f(x)=f(-x) for all x in the domain of real numbers. if you exclude the negative numbers, f(x) is one-to-one

OpenStudy (anonymous):

can u explain 0 ≤ x ≤ 6 to me? i dont think i understand how to read that

OpenStudy (anonymous):

does that mean between 0 and 6?

OpenStudy (anonymous):

less than or equal to interval notation would be [0,6]

OpenStudy (anonymous):

so the answer could only be between 0 and 6 INclusive?

OpenStudy (anonymous):

no, the inputs could only be between 0 and 6. the outputs would be the results of these inputs

OpenStudy (anonymous):

okay ill have to maul this over in my brain a bit.. thank you though

OpenStudy (anonymous):

yw

OpenStudy (anonymous):

i have an idea, find the vertex and then see whether the function is strictly increasing or decreasing over your interval, i.e. find if the vertex is in the inteval or not

OpenStudy (anonymous):

how would i go about doing that?

OpenStudy (anonymous):

since the vertex of \(y=x^4+2\) is the point (0,2) then over the interval \((0,\infty)\) your function is increasing

OpenStudy (anonymous):

and hence is one to one

OpenStudy (anonymous):

we can reason as follows: \(x^4\geq 0\) for any value of x, and so it has a minimum value at x = 0 at x = 0 the function will be decreasing on \((-\infty,0)\) and increasing on \((0,\infty)\) if the function is increasing then it is one to one, so since it is increasing on \((0,\infty)\) it is also increasing on \((0,6)\)

OpenStudy (anonymous):

so only if the function is is increasing is it one to one? and how does the given domain come into play?

OpenStudy (anonymous):

yes if it is strictly increasing (or decreasing) it is one to one

OpenStudy (anonymous):

if it is increasing this means \(f(a)<f(b)\) provided that \(a<b\) and one to one means \[f(a)=f(b)\implies a=b\] so the first statement implies the second, that is, if to inputs are different, the outputs must be different as well

OpenStudy (anonymous):

*two inputs

OpenStudy (anonymous):

this is just so complicated to me.. what is the simplist way for me to answer the question of whether it is or isnt a one to one function of an exam

OpenStudy (anonymous):

on a exam*

OpenStudy (anonymous):

how did you know that the vertix was (0,2)?

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