Let f(x) = (4x-8) / ( x^2 + 5x - 14 )
a) Find the limit as h approaches 0: ( f(1+h) - f(1) ) / h
b) Given f ' (x) = ( - 4x + 8 ) / ( (x+7) ^2 (x-2) ) For what values is f ' x NOT continuous?
c) Determine the limit of f ' (x) at each point on discontinuity found in part (b).
a) Find the limit as h approaches 0: ( f(1+h) - f(1) ) / h
b) Given f ' (x) = ( - 4x + 8 ) / ( (x+7) ^2 (x-2) ) For what values is f ' x NOT continuous?
c) Determine the limit of f ' (x) at each point on discontinuity found in part (b).
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f(x) = (4x - 8) / (x^2 + 5x - 14)
= 4(x - 2) / [(x + 7)(x - 2)] = 4 / (x +7) as long as x =/= 2
using the reduced function:
f(1 + h) = 4 / (1 + h + 7) = 4 / (h + 8)
f(1) = 4 / (1 + 7) = 4/8 = 1/2
f(1 + h) - f(1) = 4 / (h + 8) - 1/2
LCD is 2(h + 8) ==> f(1 + h) - f(1) = 8 / [2(h + 8)] - (h + 8) / [2(h + 8)]
= -h / [2(h + 8)]
[f(1 + h) - f(1)] / h = -1 / [2(h + 8)]
as h ==> 0, this goes to -1 / [2(0 + 8)] = -1/16
limit as h ==> 0 is -1/16
note that when the limit is 0 / 0, you can ususally cancel the terms that are making it go to 0 / 0, leaving a number as a limit as h ==> 0
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b) f ' (x) = (-4x + 8) / [(x + 7)^2 (x - 2)]
= -4(x - 2) / [(x + 7)^2 (x - 2)]
this is discontinuous at x = 2 and x = -7 (the zeros of the denominator)
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c) as x ==> -7, f ' (x) goes to -inf (this is a vertical asymptote, and both sides go to -inf because of the (x + 7)^2 in the denominator)
cancelling (x - 2) / (x - 2), you're left with -4 / [(x + 7)^2]
as x ==> 2, this goes to -4 / (9^2) = -4/81
limit x ==> 2 is -4 / 81
= 4(x - 2) / [(x + 7)(x - 2)] = 4 / (x +7) as long as x =/= 2
using the reduced function:
f(1 + h) = 4 / (1 + h + 7) = 4 / (h + 8)
f(1) = 4 / (1 + 7) = 4/8 = 1/2
f(1 + h) - f(1) = 4 / (h + 8) - 1/2
LCD is 2(h + 8) ==> f(1 + h) - f(1) = 8 / [2(h + 8)] - (h + 8) / [2(h + 8)]
= -h / [2(h + 8)]
[f(1 + h) - f(1)] / h = -1 / [2(h + 8)]
as h ==> 0, this goes to -1 / [2(0 + 8)] = -1/16
limit as h ==> 0 is -1/16
note that when the limit is 0 / 0, you can ususally cancel the terms that are making it go to 0 / 0, leaving a number as a limit as h ==> 0
*****
b) f ' (x) = (-4x + 8) / [(x + 7)^2 (x - 2)]
= -4(x - 2) / [(x + 7)^2 (x - 2)]
this is discontinuous at x = 2 and x = -7 (the zeros of the denominator)
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c) as x ==> -7, f ' (x) goes to -inf (this is a vertical asymptote, and both sides go to -inf because of the (x + 7)^2 in the denominator)
cancelling (x - 2) / (x - 2), you're left with -4 / [(x + 7)^2]
as x ==> 2, this goes to -4 / (9^2) = -4/81
limit x ==> 2 is -4 / 81
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NEVERMIND! I understand section (a) now! I was doing a calculation error!
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