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Solutions for Continuity

Solutions to Try Its

1. a. removable discontinuity at x=6x=6; b. jump discontinuity at x=4x=4 2. yes 3. No, the function is not continuous at x=3x=3. There exists a removable discontinuity at x=3x=3. 4. x=6x=6

Solutions to Odd-Numbered Exercises

1. Informally, if a function is continuous at x=cx=c, then there is no break in the graph of the function at f(c)f\left(c\right), and f(c)f\left(c\right) is defined. 3. discontinuous at a=3a=-3 ; f(3)f\left(-3\right) does not exist 5. removable discontinuity at a=4a=-4 ; f(4)f\left(-4\right) is not defined 7. discontinuous at a=3a=3 ; limx3f(x)=3\underset{x\to 3}{\mathrm{lim}}f\left(x\right)=3, but f(3)=6f\left(3\right)=6, which is not equal to the limit. 9. limx2f(x)\underset{x\to 2}{\mathrm{lim}}f\left(x\right) does not exist. 11. limx1f(x)=4;limx1+f(x)=1\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=4;\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=1 . Therefore, limx1f(x)\underset{x\to 1}{\mathrm{lim}}f\left(x\right) does not exist. 13. limx1f(x)=5limx1+f(x)=1\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=5\ne \underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=-1 . Thus limx1f(x)\underset{x\to 1}{\mathrm{lim}}f\left(x\right) does not exist. 15. limx3f(x)=6\underset{x\to -{3}^{-}}{\mathrm{lim}}f\left(x\right)=-6 , limx3+f(x)=13\underset{x\to -{3}^{+}}{\mathrm{lim}}f\left(x\right)=-\frac{1}{3} Therefore, limx3f(x)\underset{x\to -3}{\mathrm{lim}}f\left(x\right) does not exist. 17. f(2)f\left(2\right) is not defined. 19. f(3)f\left(-3\right) is not defined. 21. f(0)f\left(0\right) is not defined. 23. Continuous on (,)\left(-\infty ,\infty \right) 25. Continuous on (,)\left(-\infty ,\infty \right) 27. Discontinuous at x=0x=0 and x=2x=2 29. Discontinuous at x=0x=0 31. Continuous on (0,)\left(0,\infty \right) 33. Continuous on [4,)\left[4,\infty \right) 35. Continuous on (,)\left(-\infty ,\infty \right) . 37. 1, but not 2 or 3 39. 1 and 2, but not 3 41. f(0)f\left(0\right) is undefined. 43. (,0)(0,)\left(-\infty ,0\right)\cup \left(0,\infty \right) 45. At x=1x=-1, the limit does not exist. At x=1x=1, f(1)f\left(1\right) does not exist. At x=2x=2, there appears to be a vertical asymptote, and the limit does not exist. 47. x3+6x27x(x+7)(x1)\frac{{x}^{3}+6{x}^{2}-7x}{\left(x+7\right)\left(x - 1\right)} 49. fx={x2+4x12x=1fx=\begin{cases}x^{2}+4 \hfill& x\neq 1 \\ 2 \hfill& x=1\end{cases}

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  • Precalculus. Provided by: OpenStax Authored by: OpenStax College. Located at: https://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1/Preface. License: CC BY: Attribution.