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lucidriled
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 <X< 2 X^3 - (1/4)X^2 +5,X >= 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.
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
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 <x< 2 in order for this to be at least continuous, the y values of the parabola have to equal each other when x = 0. when x = 0 we get: tan-1(0) = N AND a(0)^2 + b(0) + c = N tan-1(0) = 0 AND 0 + 0 + c = 0 The only option we have for "c" then is to be 0. now that we have it at least contiuous, we need the slope of the left and right sides to approach the same value. f(x) = tan-1(x) f'(x) = 1/(x^2+1) ; at x=0, f'(0) = 1 so we know that in order to fit this missing part of the puzzle, the slope at x=0 should equate to 1 as well. f(x) = ax^2 + bx f'(x) = 2ax + b; at x=0, f'(0) = 2a(0) + b = 1 it doesnt really matter what "a" is at this point since it zeros out of the picture and we are left with: b=1 So we use what we know and fill in the missing parts: f(x) = ax^2 + 1x + 0 = ax^2 + x ; x = (0,2) If we follow the same logic at x=2, we should be able to find a suitable replacement for "a".
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 :)
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.