A function f(x) is said to have a removable discontinuity at x = a if both of the following conditions hold: 1. f is either not defined or not continuous at x = a. 2. f(a) could either be defined or redefined so that the new function is continuous at x = a. Show that f(x) = { x^2 + 8x + 19 if x < -4 0 if x = -4 -x^2 - 8x - 13 if x > -4 has a removable discontinuity at x = -4 by (a) verifying (1) in the definition above, and then (b) verifying (2) in the definition above by determining a value of f(-4) that would make f continuous at x = -4. f(-4) = would make f continuous at x = -4. Now draw a graph of f(x). It's just a couple of parabolas!
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- The function f(x) is defined as: - f(x) = x^2 + 8x + 19 if x < -4 - f(x) = 0 if x = -4 - f(x) = -x^2 - 8x - 13 if x > -4 - Check continuity at x = -4: - Left-hand limit as x approaches -4: lim (x -> -4^-) f(x) = (-4)^2 + 8(-4) + 19 = 16 - 32 + 19 Show more…
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A function f(x) is said to have a removable discontinuity at x = a if both of the following conditions hold: 1. f is either not defined or not continuous at x = a. 2. f(a) could either be defined or redefined so that the new function is continuous at x = a. Show that f(x) = x^2 + 6x + 15 if x < -3 -3 if x = -3 -x^2 - 6x - 3 if x > -3 has a removable discontinuity at x = -3 by (a) verifying (1) in the definition above, and then (b) verifying (2) in the definition above by determining a value of f(-3) that would make f continuous at x = -3. Now draw a graph of f(x). It's just a couple of parabolas!
Oswaldo J.
A function $f$ is said to have a removable discontinuity at $x=c$ if lim $_{x \rightarrow c} f(x)$ exists but $f$ is not continuous at $x=c$, either because $f$ is not defined at $c$ or because the definition for $f(c)$ differs from the value of the limit. This terminology will be needed in these exercises. (a) The terminology removable discontinuity is appropriate because a removable discontinuity of a function $f$ at $x=c$ can be "removed" by redefining the value of $f$ appropriately at $x=c .$ What value for $f(c)$ removes the discontinuity? (b) Show that the following functions have removable dis- continuities at $x=1,$ and sketch their graphs. $$ f(x)=\frac{x^{2}-1}{x-1} \quad \text { and } \quad g(x)=\left\{\begin{array}{ll}{1,} & {x>1} \\ {0,} & {x=1} \\ {1,} & {x<1}\end{array}\right. $$ (c) What values should be assigned to $f(1)$ and $g(1)$ to remove the discontinuities?
LIMITS AND CONTINUITY
Continuity
Find the values of $x$ (if any) at which $f$ is not continuous, and determine whether each such value is a removable discontinuity. $$ \begin{array}{ll}{\text { (a) } f(x)=\frac{x^{2}-4}{x^{3}-8}} & {\text { (b) } f(x)=\left\{\begin{array}{ll}{2 x-3,} & {x \leq 2} \\ {x^{2},} & {x>2}\end{array}\right.} \\ {\text { (c) } f(x)=\left\{\begin{array}{ll}{3 x^{2}+5,} & {x \neq 1} \\ {6,} & {x=1}\end{array}\right.}\end{array} $$
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