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    When f(x)=x^2+1 with a domain (-infinity, infinity), could someone explain how the range is [1, infinity) particularly how subbing x=-infinity into f(x) becomes 1
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    It doesn't. subbing "-infinity" in to it would give you infinity.

    Imagine the sketch of y = f(x). The lowest value you'd get is 1 on the y axis, because that's the value when x = 0. Either side of that, it only gets bigger and bigger infinitely. So the range is >=1
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    (Original post by bubblegum21)
    When f(x)=x^2+1 with a domain (-infinity, infinity), could someone explain how the range is [1, infinity) particularly how subbing x=-infinity into f(x) becomes 1
    it doesn't. a very negative number squared is a very large positive number. The limit of 1 comes from the case x=0
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    It's a smiley face graph, which cuts the y axis at 1.
    Draw the graph and then find the range from that graph, might be easier for u.
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    Suppose you sub in -10 for x, squaring -10 gives a positive 100 and likewise subbing in x = 10 gives 100. Any really big negative value of x will give a very big positive value which will not give 1 as you were suggesting. The minimum value that is given out by this function can only be 1 because if x = 0 then 0^2 = 0 and so 0^2 + 1 = 1.
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    I understand how it works now, my Maths teacher hasn't been very thorough! Thank you to everyone who contributed!
 
 
 
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