anonymous
  • anonymous
DOES THERE EXIST A SET OF REAL NUMBERS A, B AND C FOR WHICH THE FUNCTION TAN-1(X) X = 0 F(X)= AX^2 + BX + C, 0 = 2 IS DIFFERENTIABLE (I.E. EVERYWHERE DIFFERENTIABLE)? EXPLAIN WHY OR WHY NOT. (HERE TAN-1(X) DENOTES THE INVERSE OF THE TANGENT FUNCTION.) (From Unit 1 Exam 1 Question 5)(Sorry for the notation). According to the solutions, I don't understand how after we take the inverse of tan of 0 we equate that with C automatically. Also I don't get how at x = 0 the derivative of 5x^2 + x is 5* 0+1=1.
OCW Scholar - Single Variable Calculus
katieb
  • katieb
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anonymous
  • anonymous
First of all we know there will only be two points of discontinuity that is at x=2 or x=0 so for the function to be differentiable everywhere the L.H.D and the R.H.D at these points should be the same that is apply the following logic L.H.D at x=0 equate it to R.H.D at x=0 L.H.D at x=2 equate it to R.H.D at x=2 Use the following formula, \[\lim_{h \rightarrow 0} f(x+h)-f(x)/h\] Using that we get derivative of 5x^2 + x equal to 1 \[\lim_{h \rightarrow 0} 5(x+h)^2 + (x+h) - (5x^2+x)/h\] Solve to get answer as 1
amistre64
  • amistre64
it looks like it is asking you to "fill in" the missing piece between 0 and 2 so that the graph is at least continuous all the way across.... when we equate the left side to the missing middle we get: tan-1(x) , x=0 and ax^2 + bx + c, 0
amistre64
  • amistre64
f(x) = x^3 - (1/4)x^2 +5 f'(x) = 3x^2 -(1/2)x f(2) = (2)^3 -(2^2)/4+5 = 12 f'(2) = 3(4) -1 = 11 ....................................... so we need ..... f(2) = a(2)^2 + (2) = 12 a(4) = 10 a = 10/4 f'(2) = 2a(2) + 1 = 11 4a = 10 a = 10/4 ...................................... i got no idea where you got a=5 from ... but then maybe I forgot how to add :)

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anonymous
  • anonymous
I apologize to both but that last formula should be (5/2)X^2+X=1; as such, dhruv that is a good explanation, thank you. Amistre, thank you for the clear explanation of how C=0. My problem with the last equation was I mistakenly thought it was already in derivative form.

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