When a is strictly increasing function has a real root?
Answers
Answered by
2
If you know the derivative is positive everywhere, then you know the function has at most one root -- for example, you can reason that if it had two different roots, then the mean value theorem says that f′(x)=0f′(x)=0 for some xx between those roots, which you know isn't the case.
Unfortunately, in this case you need a little more finesse than that, because f′(1)=0f′(1)=0, as André observed. However, the argument still shows that there can't be two roots on the same side of 11 -- and you can easily calculate that f(1)>0f(1)>0, so another application of the MVT shows that there can't be a root greater than 11.
But you need to argue separately that the function has at least oneroot. The intermediate value theorem will do that for you if you show that the function has both a positive and a negative value.
Unfortunately, in this case you need a little more finesse than that, because f′(1)=0f′(1)=0, as André observed. However, the argument still shows that there can't be two roots on the same side of 11 -- and you can easily calculate that f(1)>0f(1)>0, so another application of the MVT shows that there can't be a root greater than 11.
But you need to argue separately that the function has at least oneroot. The intermediate value theorem will do that for you if you show that the function has both a positive and a negative value.
Similar questions