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Calculus/Discontinuous functions

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Expert: Alon Mandes - 9/24/2008


Question
I'm having trouble with some specifics of this problem. I think I understand the idea of continuity in functions, I'm just not sure how to apply it. The problem: f(x)= {x^2-2x-8/x-4 if x does not =4, and 3 if x=4}. I need to explain why it's discontinuous at 4.

Answer
Ok Lindsey, you can notice that the function (x^2-2x-8)/(x-4) has
singularity at x=4. Singularity means that the function will jump
to ±∞ . But let's write the function in different way :
You can confirm that (x^2-2x-8) is exact the same as (x-4)(x+2).
So now it will become : (x-4)(x+2)/(x-4) = x+2. & when x=4 f will
get the value of 4+2=6. & This is the limit of this function when
x->4. So, if we define our function as :
f(x)= {(x^2-2x-8)/(x-4) if x does not =4, and 6 if x=4}, then our
function will be continues.

Alon.

Calculus

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Alon Mandes

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Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .

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1. I'm a team member of mathnerds (math site for answering questions) 2. I'm a team member in the Student's Union of the Technion, helping students who have problems in mathematics. 3. 2 years of experience as a math teacher in college. 4. I give free homework help for high school students in Mathematics & Physics. 5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" , "Complex Functions".

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M.A in Mathematics & Bs.c in Electronics.

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