Advertisements
Advertisements
Question
Identify discontinuity for the following function as either a jump or a removable discontinuity on their respective domain:
f(x) `{:(= x^2 + 5x + 1"," , "for" 0 ≤ x ≤ 3),(= x^3 + x + 5"," , "for" 3 < x ≤ 6):}`
Advertisements
Solution
`lim_(x -> 3^-) "f"(x) = lim_(x -> 3^-) (x^2 + 5x + 1)`
= `lim_(x -> 3^-) (3)^2 + 5(3) + 1`
= 9 + 15 + 1
= 25
`lim_(x -> 3^+) "f"(x) = lim_(x -> 3^+) (x^3 + x + 5)`
= (3)3 + 3 + 5
= 35
∴ `lim_(x -> 3^-) "f"(x) ≠ lim_(x -> 3^+) "f"(x)`
∴ `lim_(x -> 3) "f"(x)` does not exist
∴ f(x) is discontinuous at x = 3
∴ f(x) has a jump discontinuity at x = 3
APPEARS IN
RELATED QUESTIONS
Examine whether the function is continuous at the points indicated against them:
f(x) = x3 − 2x + 1, for x ≤ 2
= 3x − 2, for x > 2, at x = 2
Examine whether the function is continuous at the points indicated against them:
f(x) `{:(= x^3 - 2x + 1",", "if" x ≤ 2),(= 3x - 2",", "if" x > 2):}}` at x = 2
Examine whether the function is continuous at the points indicated against them :
f(x) `{:( = (x^2 + 18x - 19)/(x - 1)",", "for" x ≠ 1),(= 20",", "for" x = 1):}}` at x = 1
Examine whether the function is continuous at the points indicated against them :
f(x) `{:(= x/(tan3x) + 2",", "for" x < 0),(= 7/3",", "for" x ≥ 0):}} "at" x = 0`
Test the continuity of the following function at the point or interval indicated against them :
f(x) `{:(= (x^3 - 8)/(sqrt(x + 2) - sqrt(3x - 2))",", "for" x ≠ 2),(= -24",", "for" x = 2):}}` at x = 2
Identify discontinuities for the following function as either a jump or a removable discontinuity :
f(x) = `(x^2 - 10x + 21)/(x - 7)`
Identify discontinuities for the following function as either a jump or a removable discontinuity :
f(x) `{:(= x^2 - 3x - 2",", "for" x < -3),(= 3 + 8x",", "for" x > -3):}`
Show that following function have continuous extension to the point where f(x) is not defined. Also find the extension :
f(x) = `(x^2 - 1)/(x^3 + 1)` for x ≠ – 1
The following function has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it become continuous :
f(x) = `((3 - 8x)/(3 - 2x))^(1/x)`, for x ≠ 0
The following function has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it become continuous :
f(x) `{:(= 3x + 2",", "for" -4 ≤ x ≤-2),(= 2x - 3";", "for" -2 < x ≤ 6):}`
The following function has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it become continuous :
f(x) `{:(= (x^3 - 8)/(x^2 - 4)",", "for" x > 2),(= 3",", "for" x = 2),(= ("e"^(3(x - 2)^2 - 1))/(2(x - 2)^2) ",", "for" x < 2):}`
If f(x) = `(sqrt(2 + sin x) - sqrt(3))/(cos^2x), "for" x ≠ pi/2`, is continuous at x = `pi/2` then find `"f"(pi/2)`
If f(x) = `(cos^2 x - sin^2 x - 1)/(sqrt(3x^2 + 1) - 1)` for x ≠ 0, is continuous at x = 0 then find f(0)
If f(x) `{:(= (5^x + 5^(-x) - 2)/(x^2)"," , "for" x ≠ 0),(= k",", "for" x = 0):}}` is continuous at x = 0, find k
Discuss the continuity of f on its domain, where f(x) `{:(= |x + 1|",", "for" -3 ≤ x ≤ 2),(= |x - 5|",", "for" 2 < x ≤ 7):}`.
Discuss the continuity of f(x) at x = `pi/4` where,
f(x) `{:(= ((sinx + cosx)^3 - 2sqrt(2))/(sin 2x - 1)",", "for" x ≠ pi/4),(= 3/sqrt(2)",", "for" x = pi/4):}`
Suppose f(x) `{:(= "p"x + 3",", "for" "a" ≤ x ≤ "b"),(= 5x^2 − "q"",", "for" "b" < x ≤ "c"):}`
Find the condition on p, q, so that f(x) is continuous on [a, c], by filling in the blanks.
f(b) = ______
`lim_(x -> "b"^+) "f"(x)` = _______
∴ pb + 3 = _______ − q
∴ p = `"_____"/"b"` is the required condition
Select the correct answer from the given alternatives:
If f(x) = `(12^x - 4^x - 3^x + 1)/(1 - cos 2x)`, for x ≠ 0 is continuous at x = 0 then the value of f(0) is ______.
Discuss the continuity of the following function at the point(s) or on the interval indicated against them:
f(x) = `(cos4x - cos9x)/(1 - cosx)`, for x ≠ 0
f(0) = `68/15`, at x = 0 on `- pi/2 ≤ x ≤ pi/2`
Identify discontinuity for the following function as either a jump or a removable discontinuity on their respective domain:
f(x) `{:(= x^2 + x - 3,"," "for" x ∈ [ -5, -2)),(= x^2 - 5,"," "for" x ∈ (-2, 5]):}`
Find k if following function is continuous at the point indicated against them:
f(x) `{:(= ((5x - 8)/(8 - 3x))^(3/(2x - 4))",", "for" x ≠ 2),(= "k"",", "for" x = 2):}}` at x = 2
Find k if following function is continuous at the point indicated against them:
f(x) `{:(= (45^x - 9^x - 5^x + 1)/(("k"^x - 1)(3^x - 1))",", "for" x ≠ 0),(= 2/3",", "for" x = 0):}}` at x = 0
Find a and b if following function is continuous at the point or on the interval indicated against them:
f(x) `{:(= (4tanx + 5sinx)/("a"^x - 1)",", "for" x < 0),(= (9)/(log2)",", "for" x = 0),(= (11x + 7x*cosx)/("b"^x - 1)",", "for" x > 0):}`
Find a and b if following function is continuous at the point or on the interval indicated against them:
f(x) `{:(= "a"x^2 + "b"x + 1",", "for" |2x - 3| ≥ 2),(= 3x + 2",", "for" 1/2 < x < 5/2):}`
Find f(a), if f is continuous at x = a where,
f(x) = `(1 - cos[7(x - pi)])/(5(x - pi)^2`, for x ≠ π at a = π
Solve using intermediate value theorem:
Show that 5x − 6x = 0 has a root in [1, 2]
Solve using intermediate value theorem:
Show that x3 − 5x2 + 3x + 6 = 0 has at least two real root between x = 1 and x = 5
If f(x) is continuous at x = 3, where
f(x) = ax + 1, for x ≤ 3
= bx + 3, for x > 3 then.
If f(x) `= sqrt(4 + "x" - 2)/"x", "x" ne 0` be continuous at x = 0, then f(0) = ______.
If f(x) = `{{:((sin5x)/(x^2 + 2x)",", x ≠ 0),(k + 1/2",", x = 0):}` is continuous at x = 0, then the value of k is ______.
If f(x) = `1/(1 - x)`, the number of points of discontinuity of f{f[f(x)]} is ______.
If the function f(x) = `[tan(π/4 + x)]^(1/x)` for x ≠ 0 is = K for x = 0 continuous at x = 0, then K = ?
For x > 0, `lim_(x rightarrow 0) ((sin x)^(1//x) + (1/x)^sinx)` is ______.
`lim_(x rightarrow 0) (e^(x^2) - cosx)/x^2` is equal to ______.
If \[\mathrm{f}(x)= \begin{cases} \mathrm{m}x+1, & x\leqslant\frac{\pi}{2} \\ \\ \mathrm{sin}x+\mathrm{n}, & x>\frac{\pi}{2} & \end{cases}\], is continuous at \[x=\frac{\pi}{2},( \begin{array} {c}\mathrm{m,n\in\mathbb{Z}} \end{array})\] then
