Question #299288

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



1
Expert's answer
2022-02-22T14:58:26-0500

1. Since "any polynomial function is continuous everywhere", f(x)=3x2+2x+1f(x) = 3x^2+2x+1 is continuous everywhere and hence continuous at x = -2


2. Since f(x)=9x21f(x)= 9x^2-1 is a polynomial, it is continuous at x = 1.


3. The function f(x)=1x2f(x)= \dfrac{1}{x-2} is discontinuous at x = 2 and it is an infinite discontinuity.


4. For the function f(x)=x1x21f(x)= \dfrac{x-1}{x^2 -1}, since both the numerator and the denominator are continuous, the discontinuities are the points obtained from x21=0 i.e., x=±1x^2-1=0\text{~i.e.,~} x = \pm1, where x=1x = 1 is a removable discontinuity since, f(x)=x1x21=1x+1f(x)= \dfrac{x-1}{x^2 -1} = \dfrac{1}{x+1}becomes continuous at x=1x = 1.


5. The function h(x)=x+1x1h(x)= \dfrac{x+1}{x-1} is discontinuous at x = 1, since limx1h(x)\displaystyle\lim_{x \rightarrow 1} h(x)\rightarrow\infty.


6. The function f(x)=x22x+1f(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)=x(x+2)(x3)f(x)= \dfrac{x}{(x+2)(x-3)} are continuous, the points of discontinuity are, (x+2)(x3)=0(x+2)(x-3)=0, i.e., x=2,3x = -2,3. Therefore, f(x)f(x) is continuous at x=2x = 2.


8. f(x)=x={x,x<0x,x0f(x)= |x| =\begin{cases}-x, &x <0\\ x, &x\geq 0\end{cases}.

We know that f(x) is continuous at x = 0 if limx0f(x)=limx0+f(x)=f(0)\displaystyle\lim _{x \rightarrow 0^{-}} f(x) = \lim _{x \rightarrow 0^{+}} f(x) = f(0).

Now,

limx0f(x)=limh0f(0h)=limh0f(h)=limh0h=limh0h=0\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}

limx0+f(x)=limh0f(0+h)=limh0f(h)=limh0h=limh0h=0\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)=0f(0) = 0. Hence f(x) is continuous at x = 0.


9. Since x24x^2 - 4 is a polynomial which is continuous everywhere, f(x)=x243f(x)= \sqrt[3]{x^2 -4} is continuous at x = 5.


10. For the function f(x)=x+1x1f(x)= \dfrac{x+1}{x-1} the point of discontinuity is x= 1, and hence f(x) is continuous at x = 0.


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