continuity

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answerhappygod
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continuity

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continuity
Continuity 1
Continuity 1 (19.64 KiB) Viewed 33 times
Continuity 2
Continuity 2 (25.52 KiB) Viewed 33 times
Continuity 3
Continuity 3 (9.88 KiB) Viewed 33 times
Continuity 4
Continuity 4 (22.47 KiB) Viewed 33 times
Continuity 5
Continuity 5 (12.54 KiB) Viewed 33 times
A function f is said to have a removable discontinuity at a if: 1. f is either not defined or not continuous at a. 2. f(a) could either be defined or redefined so that the new function is continuous at a. Let f(x)=x−22z2+5x−18​ Show that f has a removable discontinuity at 2 and determine the value for f(2) that would make f continuous at 2 . Need to redefine f(2)=
A function f(x) is said to have a jump discontinuity at x=a if: 1. x→a−lim​f(x) exists. 2. x→a+lim​f(x) exists. 3. The left and right limits are not equal. Let f(x)={6x−6,x+83​,​ if x<6 if x≥6​ Show that f(x) has a jump discontinuity at x=6 by calculating the limits from the left and right at x=6. x→6−lim​f(x)= z→5+lim​f(x)= Now, for fun, try to graph f(x).
Find the value of the constant m that makes the following function continuous on (−∞,∞). f(x)={mx−14x2+2x−7​ if x<−7 if x≥−7​ m= Now draw a graph of f.
Shown below are six statements about functions. Match each statement to one of the functions shown below which BEST matches that statement. 1. z→2lim​f(x) and x→9−lim​f(x) both exist and are fnite, but they are not equal. 2. The graph of y=f(x) has vertical tangent line at (9,f(9)) 3. z→0lim​⋅f(x)=−∞. 4. 2→+9lim​f(x) exists but t→1lim​f(x) does not. 5. c+∞lim​f(x)=∞. 6. x→0lim​f(x) exists but f is not continuous at 9 . A. f(x)=4x−9​ B. f(x)=(x−θ)1​ C. f(x)=⎩⎨⎧​cos(x−61​)02x+36​ If x<9 if x=9 if x>9​ D. f(x)=⎩⎨⎧​2x036−2z​ if x<9 if x=9 if x>9​ E. f(x)=x−91​ F. f(x)=⎩⎨⎧​2x02x−36​ if x<0 if x=9 if x>9​
f(x)={5x,x2,​x≤3x>3.​ Find the indlcated one-sided limis of f, and determine the continuly of f at the indicated point. You should also sketch a graph of y=f(x), including hollow and solid circlos in the appropriate places: NOTE: Type DNE A a limit does not exist. limx→3−​f(x)=limx→1f​f(x)=limx→3​f(x)=f(3)=​ is f continuous at 3? (YESNO)
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