Implications of the tangent of a function being at a maximum

[babayevdavid]

Does the tangent of a function being at a maximum necessarily mean that the function itself is at a maximum?

I am supposed to find whether del is at a maximum at w = (tansig*taneps)^(-1/2)

del = arctan(w*(tsig-teps)/(1+(w^2*(tsig*teps))))

tansig and taneps are constants and w is the independent variable

Using MATLAB, I've found that the derivative of tan(del) at the given w is in fact 0, and using a given graph of tan(del), I can see that the only point where the slope of the tangent is 0, is at a maximum peak. And so, I know that tan(del) at the given w is maximum.

Knowing this, can I claim that del itself is maximum at the specified w?

Thanks you all!

 

[Simon Bridge]

Does the tangent of a function being at a maximum necessarily mean that the function itself is at a maximum?

Not clear what you mean by that - the maximum possible slope for a tangent is infinity (a vertical line).

If f(x) has a maximum at x=a then the slope of the tangent at a is f'(a)=0
But that is also true if f(x) has a minimum or a point of inflexion at x=a.

Guessing that you are asking if the maxima of f(x) is also a maxima of f'(x)...
consider:

f(x)=4-x^2 has a maximum at x=0

the slope of the tangent of f at x is: f'(x) = -2x

f'(x) has a (possible) maxima when f''(x)=0 but f''(x)=-2 - so it cannot be zero!

So you see that the location of the maxima of f(x) cannot be a maxima of f'(x).

The second derivative does give you a clue though.

 

[Ray Vickson]

Does the tangent of a function being at a maximum necessarily mean that the function itself is at a maximum?

I am supposed to find whether del is at a maximum at w = (tansig*taneps)^(-1/2)

del = arctan(w*(tsig-teps)/(1+(w^2*(tsig*teps))))

tansig and taneps are constants and w is the independent variable

Using MATLAB, I've found that the derivative of tan(del) at the given w is in fact 0, and using a given graph of tan(del), I can see that the only point where the slope of the tangent is 0, is at a maximum peak. And so, I know that tan(del) at the given w is maximum.

Knowing this, can I claim that del itself is maximum at the specified w?

Thanks you all!

I'll use the symbols $b$ and $a$ instead of tsig and teps, resp., so your function is
$$f(w) = \arctan\left(\frac{(b-a)w}{1+ab\,w^2}\right)$$

If $a$ and $b$ are both > 0 then the point $w_0 = 1/\sqrt{ab}$ is, indeed, a stationary point. Whether is is a maximum or a minimum must be checked by a second-order test: $w_0$ is a maximum if $f''(w_0) < 0$ and is a minimum if $f''(w_0) > 0$, where $f''(w)$ is the second derivative of $f(w)$.

If $a$ and $b$ have opposite signs the function $f(w)$ has no stationary points, so has no points where the tangent line is horizontal. Nevertheless, it has finite maxima and minima!