^{3}- B) / (x-2)) if X<0 ,

^{2}if x>4

the problem is

let F(x) = ((x^{3} - B) / (x-2)) if X<0 ,

Ax+B if 0<x<4 ,

2x^{2} if x>4

I have to find both A ad B which make it continuous...

Tutors, sign in to answer this question.

Continuity at x=q means that f(q) is defined and lim f(x) = f(q) as x→q. This means that if the function changes at a given point (here, 0 and 4), the limit as x approaches the point of change must be the same whether x is approaching from the left or right.

Thus, as x→0 (from the left), F(x) →(-B)/(-2) = B/2

Also, as x→0 (from the right), Ax + B → B, therefore, B = B/2. B = 0 is the only value such that the limits of x as it approaches zero from the left and right are the same – the condition of continuity there.

As x → 4 (from the left), Ax + B → 4A + B = 4A

As x → 4 (from the right), 2x^{2} → 32, therefore, A = 8

The answer is A = 8 and B = 0.

As x → 4 (from the right), 2x

The answer is A = 8 and B = 0.

To make it continuous, all you need is equate consecutive formulas at their boundaries.

(x^{3} - B) / (x - 2) and Ax+B must agree at x=0.

You get (0^{3} - B) / (0 - 2) = A·0+B which is B/2 = B. This implies that B=0.

Ax+B and 2x^{2} must agree at x=4.

You get 4A+B = 32 which becomes A = (32 - B)/4 = (32 - 0)/4 = 8.

You can check now that

F(x) = (x^{3} / (x-2)) if X≤0 ,

8x if 0<x≤4 ,

2x^{2} if x>4

8x if 0<x≤4 ,

2x

is continuous in it's domain.

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