Question

Unam
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=______?? My answer is -1 but it's still says error....

Answer 1

you guys are still viewing? hehe

Answer 2

Is f(x) meant to be f(x) = 4x^3 - 3x^2 + 7x - 8÷( x-1)? What is the divisor of 8? (x-1) or just x?

Answer 3

(x-1)

Answer 4

??????????