Studying the continuity of a function?

Question

f:[0,pi] --> R f(x)=max(cos(x),cos^3(x))
x E R. How can I find out if it's continuous and derivable ?
And if it is not continuous in one point can it still be derived?

Answer 1

The functions cos and cos^3 have the same sign. Since |cos(x)| <=1 for every real x, it follows |cos^3(x)| <= |cos(x)| for every real xand, therefore,

cos(x) < 0 => cos^3(x) > cos(x)

cos(x) = 0 => cos^3(x) = cos(x)

cos(x) >0 => cos^3(x) < cos(x) This implies that

f(x) = cos(x) if x is in [0 , pi/2]
f(x) = cos^3(x) if x is in (pi/2, pi]

Since cos and cos^3 are both differentiable (the usual term for derivable), if follows f is differentiable at every point of [0 , pi] distinct from pi/2. The derivative of f to the right of pi/2 is f'+(x) = -sin(x) => f'+(pi/2) = -1, and to the left is f'-(x) = 3 cos^2(x) sin (x) => f'-(pi/2) = 0. Since the 2 dertivatives don't agree at x = pi/2, it follows f is not diffrentiable at pi/2. So, it's diffrentiable on [0, pi] - {pi/2}.

But f is continuous all over [0, pi]. Beig diffrentiable on [0, pi] - {pi/2}, f is continuous there. And at x = pi/2 it's continuous, because x -> pi/2 => cos(x) -> 0 and cos^3(x) -> 0, so the limits to the left and to the right of pi/2 are the same.

If a function is differentiable at a poit, then it's autonatically continuous there. So, if is not continuous at a given point, it can't be diffrentiable at such point. Diffentiability implies continuity, but the converse is not true.

If f and g are continuous on a set S, then max(f, g) and min(f, g) are automatically continuous at S. To see this, observe that

max(f, g) = (f + g| + |f - g| + )/2 and min(f, g) = (f - g + |f - g|)/2 .

Since f and g are continuos max(f,g) and min(f,g) are given by sums and differencesof continuous functions, being, therefore, continuous. We could have used this for your problem to concluse your f is continuous.

Answer 2

Continuity means there is no break in the function.
Derivable means there is no discontinuity in the derivitive function.

Both cosx and cos^3 x are continuous of themselves, but the question is what happens at their points of intersection.

They are equal at x = 0, pi/2, pi, 3pi/2, 2pi etc
where their values are 1, 0 , -1, 0 , 1

The derivitive functions are
-sinx
3cos^2 x * (-sinx) = -3*sinx*cos^2 x
At x = 0, both derivites = 0
at x = pi/2, -sinx = -1, the other one = 0

So the derivitives are not continuous at pi/2. Although the function has the same value, there is a jump in the derivitive.
So f(x) is continuous, but not derivable at those points.

Answer 3

If cosx is positive, then f(x) = max(cosx, cos³x) = cosx. That's for the interval [0,pi/2].

In [pi/2, pi] the cosine is negative and so f(x) = cos³x.

Obviously f is continuous and derivable in [0,pi/2) and in (pi/2, pi]. The point of contention is x=pi/2.

We have f(pi/2) = 0, and the limit is 0 whether you approach pi/2 from the left or from the right. So we have continuity.

Now for differentiability: For x f'(x) = (cosx)' = -sinx. At pi/2 this equals -1.

For x>pi/2 we have
f'(x) = (cos³x)' = 3cos²x * (-sinx). At pi/2 this is zero.

So lim(x→pi/2) f'(x) is different whether you approach it from the left or the right.

Answer: f is continuous over its entire domain.
f is differentiable over its domain except at x=pi/2.

Answer 4

max is a strange function.

It is continuous, as there is no sudden change in the value as you move through the functions domain. It is differentiable if both arguments are differentiable, although the derivative will consist of functions which cover particular ranges of the domain.

The derivative will probably not be continuous. It will be piecewise continuous over each region -- as long as each of the arguments is continuous.

Answer 5

The values of cos(x) lie between 0 & 1 (0 & 1 are included), so cos^3(x) <=cos(x). Therefore, to know the maximum between cos(x) & cos^3(x), just calculate the value of cos(x).

If you learnt cosine function, you would know it is continuous & derivable; its derivative is f(x) = -sin(x).