Answer to Question #299288 in Calculus for Klong

Question

Determine if the following functions are continuous or discontinous at the given value of x.

1. f(x)= 3x+2x+1 at x= -2

2. f(x)= 9x2-1 at x=1

3. f(x)= 1/x-2 at x=2

4. f(x)= x-1/x^2 -1 at x=1

5. h(x)= x+1/x-1at x=1

6. f(x)= x^2 -2x+1 at x=3

7. f(x)= x/(x+2)(x-3) at x=2

8. f(x)= |x| at x=0

9. f(x) Cube root√x^2 -4 at x=5

10. f(x)= x+1/x-1 at x=0

Answer 1

1. Since "any polynomial function is continuous everywhere", $f(x) = 3x^2+2x+1$ is continuous everywhere and hence continuous at x = -2

2. Since $f(x)= 9x^2-1$ is a polynomial, it is continuous at x = 1.

3. The function $f(x)= \dfrac{1}{x-2}$ is discontinuous at x = 2 and it is an infinite discontinuity.

4. For the function $f(x)= \dfrac{x-1}{x^2 -1}$, since both the numerator and the denominator are continuous, the discontinuities are the points obtained from $x^2-1=0\text{~i.e.,~} x = \pm1$, where $x = 1$ is a removable discontinuity since, $f(x)= \dfrac{x-1}{x^2 -1} = \dfrac{1}{x+1}$becomes continuous at $x = 1$.

5. The function $h(x)= \dfrac{x+1}{x-1}$ is discontinuous at x = 1, since $\displaystyle\lim_{x \rightarrow 1} h(x)\rightarrow\infty$.

6. The function $f(x)= x^2 -2x+1$ is continuous at x=3 as it is a polynomial function.

7. Since both numerator and denominator of the function $f(x)= \dfrac{x}{(x+2)(x-3)}$ are continuous, the points of discontinuity are, $(x+2)(x-3)=0$, i.e., $x = -2,3$. Therefore, $f(x)$ is continuous at $x = 2$.

8. $f(x)= |x| =\begin{cases}-x, &x <0\\ x, &x\geq 0\end{cases}$.

We know that f(x) is continuous at x = 0 if $\displaystyle\lim _{x \rightarrow 0^{-}} f(x) = \lim _{x \rightarrow 0^{+}} f(x) = f(0)$.

Now,

$\begin{aligned} \lim _{x \rightarrow 0^{-}} f(x) &=\lim _{h \rightarrow 0} f(0-h) \\ &=\lim _{h \rightarrow 0} f(-h) \\ &=\lim _{h \rightarrow 0}|-h| \\ &=\lim _{h \rightarrow 0} h \\ &=0 \end{aligned}$

$\begin{aligned} \lim _{x \rightarrow 0^{+}} f(x) &=\lim _{h \rightarrow 0} f(0+h) \\ &=\lim _{h \rightarrow 0} f(h) \\ &=\lim _{h \rightarrow 0}|h| \\ &=\lim _{h \rightarrow 0} h \\ &=0 \end{aligned}$

and $f(0) = 0$. Hence f(x) is continuous at x = 0.

9. Since $x^2 - 4$ is a polynomial which is continuous everywhere, $f(x)= \sqrt[3]{x^2 -4}$ is continuous at x = 5.

10. For the function $f(x)= \dfrac{x+1}{x-1}$ the point of discontinuity is x= 1, and hence f(x) is continuous at x = 0.