Mean Value Theorem

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Jason76

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\(\displaystyle f(x) = \sin(x)\)

\(\displaystyle g(x) = \cos(x)\)

For what value of \(\displaystyle a\) is the line tangent to the graph of \(\displaystyle f\) at \(\displaystyle x = a\) parallel to the line tangent to the graph of \(\displaystyle g\) at \(\displaystyle x = a?\).

I think this is mean value theorem so

Given \(\displaystyle x\) interval of \(\displaystyle [\cos(x), \sin(x)]\)

\(\displaystyle \dfrac{f[\sin(x)] - f[\cos(x)]}{\sin(x) - \cos(x)}\)

or could this be something parametric, or something else?
 
Last edited:
Jason76 said:
\(\displaystyle f(x) = \sin(x)\)

\(\displaystyle g(x) = \cos(x)\)

For what value of \(\displaystyle a\) is the line tangent to the graph of \(\displaystyle f\) at \(\displaystyle x = a\) parallel to the line tangent to the graph of \(\displaystyle g\) at \(\displaystyle x = a?\)
Click to expand...
Jason76, was there an interval given for where x = a lies?
 
You are told that f(x)= sin(x) and g(x)= cos(x). So why in the world are you talking about "f[sin(x)]" and "f[cos(x)]". And why are you ignoring "g"?

No, this problem has nothing to do with the "mean value theorem". it has everything to do with knowing basic concepts- two lines are parallel if and only if they have the same slopes and the slope of a tangent line is the derivative of the function. So this problem is asking you to find x such that f'(x)= g'(x). That is, so that (sin(x))'= (cos(x))' where the ' means the derivative. What are the derivatives of sine and cosine?
 
HallsofIvy said:
...- two lines are parallel if and only if they have the same slopes...
Click to expand...


That is partially incorrect.

Both lines having the same slopes or both lines having no slopes (undefined slopes) are necessary

conditions, but they are not sufficient conditions. Those conditions do not keep

the lines from being coincident, which are not parallel to each other.
 
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