Question

Find the Value of k so that the function f is continuous:

f(x)= {9x^2-4/3x-2, x≠-2/3 }

k , x=-2/3

Answer 1

Answer:

$k=\frac{26}{9}$

Step-by-step explanation:

$f(x)=9x^2-\frac{4}{3}x-2, x\neq \frac{-2}{3}$ and $f(x)=k, x=\frac{-2}{3}$

Remember that a function is continuous in x=a if $lim_{x\rightarrow a}f(x)=f(a)$

Since f is a rational function for every number different of $x=-\frac{2}{3}$, then f is continuous in these numbers.

Then we need that

$lim_{x\rightarrow\frac{-2}{3}}f(x)=f(\frac{-2}{3})=k$

As we are approaching -2/3 then we are taking values greater and less than -2/3. Therefore for the limit we use $f(x)=9x^2-\frac{4}{3}x-2$.

But observe that $9(\frac{-2}{3})^2-\frac{4}{3}\frac{-2}{3}-2=\frac{26}{9}$.

Then k must be $\frac{26}{9}$.

let's verify that in effect f is continuous with $\frac{26}{9}$.

$lim_{x\rightarrow \frac{-2}{3}}f(x)=lim_{x\rightarrow \frac{-2}{3}} (9x^2-\frac{4}{3}x-2)=9(\frac{-2}{3})^2-\frac{4}{3}*\frac{-2}{3}-2=\frac{26}{9}=k=f(\frac{-2}{3})$