$$f(x) = \frac{2x^2+2x-40}{x-4}$$

Show that $f$ has a removable discontinuity at $4$ and determine the value for $f(4)$ that would make $f$ continuous at $4$.

I know what a removable discontinuity is but I don't quite understand how I could make a value for $f(4)$ that would lead to $f$ being continuous. Especially when $x = 4$ is undefined due to the denominator being zero.

Any help?

• Divide the numerator by $x-4$. Can you divide polynomials? – wjm Oct 3 '17 at 4:08
• Yes. I did what you said and got 2x + 10. Is this correct? If so, what am I supposed to do after? Thank you! – sktsasus Oct 3 '17 at 4:14
• Use the definition of continuity of a function at a point – Anil Kumar Pandey Oct 3 '17 at 4:31
• For your function to be continuous f(4) must be equal to the limit of the function at x=4 – Anil Kumar Pandey Oct 3 '17 at 4:33
• Now calculate the limit of the function at x=4 – Anil Kumar Pandey Oct 3 '17 at 4:34

f(x)=(2x2+2x−40x−4)/x-4. ---------- 1

2x^2+2x-40

2(x^2+x-20) factorizing to 2(x+5)(x-4) ---------- 2.

Substituting 2 in 1 gives.

2(x+5)(x-4)/(x-4)

equals to 2(x+5)

therefore it has a removable discontinuity at x=4.