anonymous
  • anonymous
how would i figure out if f(x) = x4 + 2; 0 ≤ x ≤ 6 is a "one to one" function?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
you could find the inverse, if the inverse if a function over the domain. then f(x) is one-to-one
anonymous
  • anonymous
the domain being 0 ≤ x ≤ 6 ?
anonymous
  • anonymous
yes

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More answers

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

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