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Grace [21]
3 years ago
15

Can someone please help me with continuity? I would like to check my work, I’m a bit confused on the differentiability part

Mathematics
1 answer:
nordsb [41]3 years ago
6 0

For <em>f</em> to be <u>continuous</u> at some point <em>x</em> = <em>c</em>, you require that

• <em>f(c)</em> exists and is finite, and

• the limits of <em>f</em> as <em>x</em> approaches <em>c</em> from either side match, and their value must be <em>f(c)</em>

<em />

For <em>f</em> to be <u>differentiable</u> at <em>x</em> = <em>c</em>, you require that

• <em>f</em> is continuous at <em>c</em> (i.e. the above conditions are all met), and

• the derivative <em>f '</em> is continuous at <em>c</em>

Without having the graph of <em>f '</em>, you can assess whether the last condition is met by some function by mentally tracing a tangent line to the graph as you get closer to <em>c</em>. If the slope of the tangent doesn't change, then the function is differentiable. This is usually accompanied by jumps or sharp corners in the graph.

<em />

Some examples:

• every polynomial is both continuous and differentiable

• the absolute value function |<em>x</em>| is continuous but not differentiable at <em>x</em> = 0

• 1/<em>x</em> is neither continuous nor differentiable at <em>x</em> = 0

(a) Neither continuous nor differentiable

Why? <em>f</em> has a vertical asymptote at <em>x</em> = -2 and <em>f</em> (-2) does not exist. <em>f</em> is not continuous, and therefore not differentiable.

(b) Neither continuous nor differentiable

Why? From the left, <em>f</em> is approaching 2.5, while from the right, it's approaching 5.

(c) Neither continuous nor differentiable

Why? The limits from either side of <em>x</em> = 2 match and are equal to 0.5, but <em>f</em> (2) itself does not exist (which is indicated by the point being a hollow circle).

If the circle was instead filled in, so that <em>f</em> (2) = 0.5, then <em>f</em> would be continuous there, but still not differentiable. Notice the sharp corner. Or, using the tangent-line analysis, the slope of the tangent to the left of <em>x</em> = 2 is negative, but to the right it would be positive.

(d) Both continuous and differentiable

Why? <em>f</em> doesn't have any special features at <em>x</em> = 3 that would suggest it's not continuous nor differentiable.

(e) Neither continuous nor differentiable

Why? Similar reasoning as in (a). There's another vertical asymptote, but the graph shows <em>f</em> (4) = 1. However, the limits from either side of <em>x</em> = 4 are positive infinity, not 1.

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