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
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.
I agree with Jim
I also got 1/(4√13) for the slope
To find the derivative of a function, we need to use the rules of differentiation. In this case, we are given the function f(x) = √(x + 6) and we want to find the derivative at x = 46.
The first step is to find the derivative of f(x) with respect to x, denoted as f'(x) or dy/dx. We can use the power rule and chain rule to find the derivative:
f'(x) = (1/2) * (x + 6)^(-1/2) * 1
Now, we want to find the slope of the tangent line at x = 46, which means we need to substitute x = 46 into the derivative. Let's calculate it:
f'(46) = (1/2) * (46 + 6)^(-1/2) * 1
= (1/2) * (52)^(-1/2) * 1
Simplifying further gives:
f'(46) = (1/2) * 1/√52
From here, we can see that the expression is in the form of 1/n, where n is an integer. To determine n, we need to rationalize the denominator (√52) by multiplying both the numerator and denominator by √52:
f'(46) = (1/2) * 1/√52
= (1/2) * (1/√52) * (√52/√52)
= (1/2) * √52/52
= √52/104
Comparing this result with the given statement that the derivative is 1/n, we can identify that n = 104. Therefore, in this case, we expect n to be 104.