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you could find the inverse, if the inverse if a function over the domain. then f(x) is one-to-one
the domain being 0 ≤ x ≤ 6 ?
if f(x) outputs only 1 value for each x in the domain, then f(x) is one-to-one
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
can u explain 0 ≤ x ≤ 6 to me? i dont think i understand how to read that
does that mean between 0 and 6?
less than or equal to interval notation would be [0,6]
so the answer could only be between 0 and 6 INclusive?
no, the inputs could only be between 0 and 6. the outputs would be the results of these inputs
okay ill have to maul this over in my brain a bit.. thank you though
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
how would i go about doing that?
since the vertex of \(y=x^4+2\) is the point (0,2) then over the interval \((0,\infty)\) your function is increasing
and hence is one to one
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 \((0,6)\)
so only if the function is is increasing is it one to one? and how does the given domain come into play?
yes if it is strictly increasing (or decreasing) it is one to one
if it is increasing this means \(f(a)
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
on a exam*
how did you know that the vertix was (0,2)?