Question

the function f is given by the formula

ƒ(x)= 4x^3-3x^2+7x-8÷ x-1

when x<1 and by the formula
ƒ(x)=-4x^2-2x+a
when 1≤x

What value must be chosen for a in order to make this function continuous at 1?

a=______??

Answer 1

For f(x) to be continuous at x=1
left hand limit of f at x=1 should be equal to the right hand limit at x=1 equal to f(1)
f(1) = 4-2+a = 2+a
left hand limit of f at x=1 is f(1-h) = \[\lim_{h \rightarrow 0}\]( 4(1-h)^3-3(1-h)^2+7(1-h)-8)/h
since h tends to zero so we can use binomial expansion
(1-h)^n = 1-nh-n*n-1*h*h/2........
Neglecting higher powers...
f(1-h) = (4(1-3h)-3(1-2h)+7+7h-8)/h
f(1-h) = (4-12h-3+6h+7+7h-8)/h = h/h =1
f(1) = 2+a = f(1-h) = 1
2+a = 1
Thus a= -1

Answer 2

Hi Suyash...i got the same answer as well but i still got it wrong when i plugged it in :)