Question

Suppose f(x)=A+x for x≺2 and f(x)=3+x^2 for x≥2 . Find a value of A such that the function f(x) is continuous at the point x=2 .

I got -5 as an answer but it's not right so I am lost as to what the answer is. What I did is I added 2 into the equation 3+x^2 and I got seven then I put A+x and set it as A+7=2 and that is how I got -5. I have a feeling I am doing this way wrong.

Answer 1

So we have:
\[f(x)=\left\{\eqalign{
&A+x, \phantom{.}x\lt2 \\
&3+x^2, \phantom{.}x\geq2 \\
}\right\}\]
Right?
And we need it to be countinuous at \(x=2\) right?

Answer 2

correct!

Answer 3

Alright. Well for the function \(f(x)\) to be continuous, we need to have the following equation defined:
If \(f(x)\) is to be continuous at \(x=a\), then:
\[\lim_{x\rightarrow a^-}{f(x)}=\lim_{x\rightarrow a^+}{f(x)}\]
So let us apply that condition here:
\[\lim_{x\rightarrow 2^-}{[f(x)]_1}=\lim_{x\rightarrow 2^+}{[f(x)]_2}\]
I have subscripted both instances of the function because, co-incidentally, at that point they are two different functions depending on which side you approach 2 from. Good so far?

Answer 4

So we can make the substitution \(\bigg([f(x)]_1\rightarrow A+x\bigg)\) and \(\bigg([f(x)]_2\rightarrow 3+x^2\bigg)\)
Then we can simplify:
\[\eqalign{
&\lim_{x\rightarrow 2^-}{A+x}=\lim_{x\rightarrow 2^+}{3+x^2} \\
&\lim_{x\rightarrow 2}{A+x}=\lim_{x\rightarrow 2}{3+x^2} \\
&\lim_{x\rightarrow 2}{A+x}=7 \\
&A+2=7 \\
&A=7-2 \\
&A=5 \\
}\]
Seems that the absolute value of your answer was correct, but not the sign haha cool?

Answer 5

Oh okay makes sense! Thank you so much!

Answer 6

No prob orbie! My pleasure