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If the function f(x) is differentiable and f(x) = {(ax^3)-6x at x<1 (and equal to) and (x^2) + 4 at x >1} then a=? show all steps?

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  • If the function f(x) is differentiable and f(x) = {(ax^3)-6x at x<1 (and equal to) and (x^2) + 4 at x >1} then a=? show all steps?


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Answer #3 | 19/11 2016 07:01
f(x) is differentiable. For x( 1, f(x) =x^2. Since f(x) is differentiable, f(x) must be continuous at x = 1. Then a(1)^3 -6(1) = (1)^1, ie., a-6 = 1, ie., a =7.
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Answer #13 | 18/11 2016 21:47
If it has one formula above 1, and a different formula below 1, but it's still differentiable, then both formulas must equal the same value at x=1. So: (a*1^3) - 6*1 = (1^2) + 4 (a*1^3) - 6*1 = 5 a - 6 = 5 a = 11
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Answer #14 | 19/11 2016 05:47
If it has one formula above 1, and a different formula below 1, but it's still differentiable, then both formulas must equal the same value at x=1. So: (a*1^3) - 6*1 = (1^2) + 4 (a*1^3) - 6*1 = 5 a - 6 = 5 a = 11
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Answer #15 | 19/11 2016 06:06
. should be continuous at x = 1 ax³ - 6x = x² ax³ - x² - 6x = 0 x(ax² - x - 6) = 0 x = 0 and ax² - x - 6 = 0 ax² - x - 6 = 0 a = (x + 6)/x² a = (1 + 6)/1² a = 7 ———
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Answer #16 | 25/11 2016 08:41
If f is differentiable at x=1, then it is continuous at x=1 f(1-) = lim x-->1- f(x) = lim x-->1- (ax^3-6x) = a-6 f(1+) = lim x-->1+ f(x) = lim x-->1+ (x^2+4) = 5 In order for the limit to exist, f(1+)= f(1-) a-6 = 5 a = 11 Also, you should define f(1) = 5 In your question f is not defined when x=1 You may change it as ax^3-6x at x <=1 or x^2+4 at x >= 1
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Answer #17 | 18/11 2016 23:01
f(x) is differentiable. For x( 1, f(x) =x^2. Since f(x) is differentiable, f(x) must be continuous at x = 1. Then a(1)^3 -6(1) = (1)^1, ie., a-6 = 1, ie., a =7.
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