Asked by BB22
The function f(x)= sqrt(x+6), I am told that the derivative (slope of tangent line) at x=46 is 1/n for some integer n. what do I expect n to be? I don't understand this question very well. Help is greatly appreciated
Answers
Answered by
jim
I can't find an integer n for which this is true.
Consider f'(x) = 1/(2sqrt(x+6))
=1/2sqrt(52) at point x=46
You can take a factor of 4 out of that,
=1/4sqrt(13)
but you're still left with the irrational sqrt(13). You can't get an integer out of it unless x+6 is a square.
Unless there's a typo, or I'vs slipped a cog, I think there may be a problem with your question.
Consider f'(x) = 1/(2sqrt(x+6))
=1/2sqrt(52) at point x=46
You can take a factor of 4 out of that,
=1/4sqrt(13)
but you're still left with the irrational sqrt(13). You can't get an integer out of it unless x+6 is a square.
Unless there's a typo, or I'vs slipped a cog, I think there may be a problem with your question.
Answered by
Reiny
I agree with Jim
I also got 1/(4√13) for the slope
I also got 1/(4√13) for the slope
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