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Nataliya [291]
3 years ago
9

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If a fu

nction is continuous at a point, then it is differentiable at that point.
Mathematics
1 answer:
natta225 [31]3 years ago
5 0

Answer:

See Explanation

Step-by-step explanation:

If a Function is differentiable at a point c, it is also continuous at that point.

but be careful, to not assume that the inverse statement is true if a fuction is Continuous it doest not mean it is necessarily differentiable, it must satisfy the two conditions.

  • the function must have one and only one tangent at x=c
  • the fore mentioned tangent cannot be a vertical line.

                                          And

If function is differentiable at a point x, then function must also be continuous at x. but The converse does not hold, a continuous function need not be differentiable.

  • For example, a function with a bend, cusp, or vertical tangent may be continuous, but fails to be differentiable at the location of the anomaly.

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The dryers at Quik-Wash Laundry cost a quarter for 7 minutes. How many quarters will you have to use to dry your clothes for 45
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Give the number to which the Fourier series converges at a point of discontinuity of f.
quester [9]

Answer:

The Fourier series of f(x) converges to 3 at the points x= π+2kπ, where k is an integer.

Step-by-step explanation:

First, recall that the function f(x) is extended 2π periodic to the whole real line, in order to obtain a valid Fourier expansion. Remember that a Fourier series is formed by a sines and cosines, which are 2π-periodic.

So, the 2π-periodic expansion of f(x) is discontinuous at the points π+2kπ, in particular π and -π. Check the attached figure to a better understanding.

Now, the Dirichlet theorem on the convergence of a Fourier series tells us that the series converges to the function at the points of continuity, and at points of discontinuity the sum of the series is

\frac{f(x_0+)+f(x_0-)}{2}.

Here we understand the notation f(x+) and f(x-) as

f(x_0+) = \lim_{x\rightarrow x_0}f(x), x>x_0

and

f(x_0-) = \lim_{x\rightarrow x_0}f(x), x.

In this particular case

f(\pi-) = \lim_{x\rightarrow \pi}(3-2x) = 3-2\pi, x.

For the limit f(\pi+) = \lim_{x\rightarrow \pi}(3-2x), with x>\pi recall that our function is 2π-periodic, so the values of f near π, with x>π are the same when x is near -π and x>-π. Again, check the attached figure. So,

f(\pi+) = \lim_{x\rightarrow \pi}(3-2x) = 3+2\pi, x.

Thus,

\frac{f(\pi+)+f(\pi-)}{2} = \frac{3+2\pi +3-2\pi}{2} = \frac{6}{2} =3.

Note: In the attached figure we only have drawn three repetitions of the 2π-periodic extension of f, recall that the extension is <em>ad infinitum</em>. Also, the points drawn in the dotted lines are the sum of the series at the points of discontinuity.

6 0
3 years ago
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