TheJoe6 From: US Posts: 36 Votes: 144 |
Determine if f is differentiable at x=a 5. f(x) = {-2x^2 x< 0; 2x^2 x=> 0 a = 0 6. f(x) = {x^2 - 5 x< 3; 3x - 5 x=> 3 a = 3 7. f(x) = { Square root(2-x) x<2; (2-x)^2 x=>2 a=2 |

Anonymous From: - Posts: - Votes: - |
This famous result by Euler is rather straightforward to prove using Taylor Series expansions, which I hop you know a bit about. So we are trying to show: (1 + a/x)^x -> e^a Well take logartihms of both sides to make things simpler: We are trying instead to show: x*log(1 + a/x) -> a as x -> inf Now, x tending to infinity isn't very pleasant to deal with, we are better equipped with things tending to zero. So let us replace x by 1/y: We are now trying to show: 1/y * log(1 + ay) -> a as y -> 0 Okay, now you need to use the taylor series expansion for log. It is one of the most important expansions, so you might need to know it depending on your level of maths, but look it up on wiki if you don't. Here is the standard expansion: log(1 - z) = - ( z^n / n summed from n = 1 to n = inf). Remember this is only true for -1 <= z < 1 (which is why we had to substitute the y in to get small numbers). So if we use the expansion with our expression, we are left with: ( (-1)^(n+1) * a^n * y^n-1 / n summed from n = 1 to n = inf ) What is this sum when y gets very small? Well for n>1 the terms have a factor of y in them, so they tend to zero, so ignore them. The only big term is for n = 1, which is a. So in fact our expression tends to a, which is what we expected. NEXT f(x) = x^2 - 2 when x not 3 and 5 when x = 3 You are correct that x^2 is continuous everywhere, however we are not trying to show x^2 has a discontinuity, we are trying to show f has a discontinuity, i.e. does the graph of f 'jump' somewhere? Well for x not 3, clearly it is a nice smooth continuous line on the graph, so we are trying to show it jumps at exactly x = 3. Do you remember the definition of f being continuous at a point a? It means, as x tends to a, f(x) tends to f(a). Is this true for a = 3? As x tends to 3, we know that x^2 - 2 is continuous so it tends 3^2 - 2 = 7. So f(x) tends to 7, but this is not f(3)! f(3) is actually 5, so the graph actually jumps down to 5 and jumps back up to 7 as x increases; a clear discontinuity. NEXT Okay, remember your definition. g(x) is differentiable at 1 if: lim{h -> 0} ((g(1 + h) - g(1)) / h ) exists well, what is this expression? It differs whether h is positive or negative. Let's assume for now h is negative. So the expression is: lim{h -> 0} (1 - h - 1)/h = lim{h -> 0} -1 = -1 which certainly exists. Now let's try for h positive: I'll leave you to expand the algebra, but you get the limit to be 0. So the limits exist either side, BUT they are not the same. Hence the actual limit does not exist, so g is not differentiable. Note that I went the long way about this question, because it is the general method you should use. However it is easier to spot that both parts of the g function (2-x) and (x^2 - 2x + 2) are well-known differentiable functions, and their derivatives are -1 and 0 at the point x = 3 respectively. Since the derivatives aren't equal, this means the composition g is not a differentiable function. |

Anonymous From: - Posts: - Votes: - |
<p>(1)</p> <p>(1 + a/x)^x -> e^a<br /> x*log(1 + a/x) -> a as x -> inf<br /> So let us replace x by 1/y:<br /> 1/y * log(1 + ay) -> a as y -> 0<br /><br /> log(1 - z) = - ( z^n / n summed from n = 1 to n = inf). Remember this is only true for -1 <= z < 1 <br />So if we use the expansion with our expression, we are left with:<br />( (-1)^(n+1) * a^n * y^n-1 / n summed from n = 1 to n = inf )<br />The only big term is for n = 1, which is a.<br /><br /><br />(2)<br /><br />f(x) = x^2 - 2 when x not 3<br />and 5 when x = 3<br />You are correct that x^2 is continuous everywhere, however we are not trying to show x^2 has a discontinuity, we are trying to show f has a discontinuity, Well for x not 3, clearly it is a nice smooth continuous line on the graph, so we are trying to show it jumps at exactly x = 3.<br />as x tends to a, f(x) tends to f(a).</p> <p>Is this true for a = 3? As x tends to 3, we know that x^2 - 2 is continuous so it tends 3^2 - 2 = 7. So f(x) tends to 7, but this is not f(3)! f(3) is actually 5, so the graph actually jumps down to 5 and jumps back up to 7 as x increases; a clear discontinuity.<br /><br />(3)<br /><br />lim{h -> 0} ((g(1 + h) - g(1)) / h ) exists<br />lim{h -> 0} (1 - h - 1)/h = lim{h -> 0} -1 = -1 which certainly exists.<br />I'll leave you to expand the algebra, but you get the limit to be 0.<br />So the limits exist either side, BUT they are not the same. Hence the actual limit does not exist, so g is not differentiable.</p> |

Anonymous From: - Posts: - Votes: - |
(1) (1 + a/x)^x -> e^a x*log(1 + a/x) -> a as x -> inf So let us replace x by 1/y: 1/y * log(1 + ay) -> a as y -> 0 log(1 - z) = - ( z^n / n summed from n = 1 to n = inf). Remember this is only true for -1 <= z < 1 So if we use the expansion with our expression, we are left with: ( (-1)^(n+1) * a^n * y^n-1 / n summed from n = 1 to n = inf ) The only big term is for n = 1, which is a. (2) f(x) = x^2 - 2 when x not 3 and 5 when x = 3 You are correct that x^2 is continuous everywhere, however we are not trying to show x^2 has a discontinuity, we are trying to show f has a discontinuity, Well for x not 3, clearly it is a nice smooth continuous line on the graph, so we are trying to show it jumps at exactly x = 3. as x tends to a, f(x) tends to f(a). Is this true for a = 3? As x tends to 3, we know that x^2 - 2 is continuous so it tends 3^2 - 2 = 7. So f(x) tends to 7, but this is not f(3)! f(3) is actually 5, the graph actually jumps down to 5 and jumps back up to 7 as x increases; a clear discontinuity. (3) lim{h -> 0} ((g(1 + h) - g(1)) / h ) exists lim{h -> 0} (1 - h - 1)/h = lim{h -> 0} -1 = -1 which certainly exists. I'll leave you to expand the algebra, but you get the limit to be 0. So the limits exist either side, BUT they are not the same. Hence the actual limit does not exist, so g is not differentiable. |

iSam15 From: US Posts: 9 Votes: 27 |
A. f(x) = {-2x^2 x< 0; 2x^2 x=> 0 f' = -4x 4x a = 0 LHL = 0=RHL =>CONTINUOUS & Differentiable. b. f(x) = {x^2 - 5 x< 3; 3x - 5 x=> 3 a = 3 LHL = 4 RHL = 1 => Not CONTINUOUS & hence Differentiable. c. f(x) = { Square root(2-x) x<2; (2-x)^2 x=>2 a=2 LHL= 0=RHL => LIMIT EXISTS. f' = -0.5 /(2-x) for x<2 -2(2-x) for x> 2 For x<2 limit f' -> ∞ ∴Not CONTINUOUS & hence Differentiable |

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