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Homework Statement
Show f(x)=(x−2)sinxln(x+2) has f'(x)=0 somewhere on [-1,3]
The Attempt at a Solution
I tried using Rolle's theorem, but f(-1)≠f(3). Then I tried the mean value theorem, but didn't get 0 either.
You can find f'(x) analytically, so maybe you should do that and then apply the intermediate value theorem.Homework Statement
Show f(x)=(x−2)sinxln(x+2) has f'(x)=0 somewhere on [-1,3]
The Attempt at a Solution
I tried using Rolle's theorem, but f(-1)≠f(3). Then I tried the mean value theorem, but didn't get 0 either.
Well if it's numerical analysis, why are you using those theorems? Suppose it was some function that you didn't even know what the form was, how could you show the derivative is zero somewhere in that interval? If it was mine, I'd generate "numerically" since this is numerical analysis, a set of equally spaced points in that interval and then compute the value of the function and then I'd inspect the list for the "monotonicity" change, that is, when the numbers are increasing then decreasing or vice-a-versa. However if there is an inflection point, the derivative could still be zero without this change in monotonicity. But we could at least rule out the former case this way. Anyway, lots of tough problems won't give you the luxury of using nice theorems and you'll have to muscle-through the data in this way or another.Homework Statement
Show f(x)=(x−2)sinxln(x+2) has f'(x)=0 somewhere on [-1,3]
The Attempt at a Solution
I tried using Rolle's theorem, but f(-1)≠f(3). Then I tried the mean value theorem, but didn't get 0 either.
What does the interval [0,1] do for you?I got it, make [0,1] a subset of [-1,3] then i can use Rolle's theorem and then determine that there must be a c in [0,1] so there's gotta be a c in [-1,3]
sorry i meant for the interval to be [0,2]What does the interval [0,1] do for you?
That's better.sorry i meant for the interval to be [0,2]
then i get f(0)=f(2)=0 => rolle's