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Answer:

Let's go from the top to bottom on this one.

(i) If `f` is continuous at `x=c`, then `f'(c)` exists.

Well, let's look at what it means for `f(x)` to be continuous and differentiable. The definition for continuity is given by:

`lim_(x->c) f(x) = f(c)`

So, we know that that limit holds true for the domain of the function. However, if we look at the conditions for differentiability, we see two: that `f` is continuous, and that `f` is "smooth." Another way to think of this, the derivative of `f` must be continuous at `x=c`.

So, in order to show that (i) is not true, we just need to find a counterexample. In other words, we need a continuous function that is not differentiable at all points. This function comes to mind:

`f(x) = |x|` (absolute value function)

If we plot this function we get:

Now, maybe you can see that the graph isn't "smooth" at x = 0. Well, let's show the derivative to make clear the discontinuity:

The derivative clearly does not exist at `f(0)`. Therefore, (i) must be **false**.

(ii) If `f'(c)` exists, then `f` is continuous at `x=c`.

Well, if you look at our answer for the previous question, we stated explicity that in order for the derivative to exist, the function must be both continuous and smooth at a certain point.

Therefore, if `f` is differentiable at `x=c` then it must be continuous!

(ii) must be **true**.

(iii) `lim_(x->c)f(x) = f(c)`

Well, this, as we discussed above, is a definition for continuity. However, we made no stipulation other than that `f(x)` is a function. Not every function is continuous, therefore (iii) must be **false**. Well, true sometimes, but in math, we say false when it's true sometimes, like in (i)!

(iv) if `f` is continuous on `(a,b)` then `f` is continuous on `[a,b]`.

Well, this is tricky if you mix up your interval notation! Remember, parantheses indicates an open interval. This means that `f` being continuous does not necessarily apply to `a` or `b`! So, when we put the closed brackets on the interval, we include `a` and `b`! We don't know what's going on at those points based on the premise!

Here's an example: `f(x) = 1/x`

` `Now, don't get worried about the rational/hyperbolic function, it's just there to illustrate a point. Let's say our open interval is `(0, 1)`. Well, we know we can equate the limit from the left and right for any point **between **0 and 1; however, can we do the same at 0 and 1?

Let's look at the graph:

Well, it certainly looks continuous at `x=1` but what's going on at 0? Does that line ever cross the y-axis?

Well, if we evaluate `f(x) = 1/x` you'll see that we'd divide by 0 if `x=0`

` `Another way to see that the function is not continuous is to see the graph of the negative side, too:

Even if the graph on the right crossed the y-axis, clearly the limit from the right is NOT the limit from the left! Therefore

`lim_(x->0) 1/x` does not exist!

So, two things can happen when you go from an open interval to a closed one. You may include a point that is not defined on the function, or you may include a point whose limit does not exist.

Either way, (iv) is **false**.

Hope that helps!

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