1.Which function is NOT continuous everywhere?
A.y=(4x)/(x+1)^2
B.y=|x|
C.y=(x)/(x^2+1)
D.y=√(x^2+1)
E.y=x^(2/3)
2.State the domain f(x)= √((x+6)) sinx
3. Let h(x) {(sin2x)/x, if x ≠ 0
{a, if x=0
find the value of a so that hi s continuousat x=0
a. it is sot possible to find the value of a so that h(x) is cont. at x=0
b. a=1
c.a=1/2
e.a=0
If the function f is cont. for all positive real numbers and if f(x)=(lnx^2-xlnx)/(x-2) when x ≠ 2, the f(2)=?
Clearly A is not continuous at x = -1, since the denominator is zero.
The domain of √u is u >= 0.
So, we need x+6 >= 0
The domain of sin x is all reals, so it doesn't affect the result.
we know that lim sin(x)/x = 1, so
sin(2x)/(2x) is 1. So, lim sin(2x)/x = 2.
ln x^2 = 2lnx, so we have
(2lnx - xlnx)/(x-2) = (2-x)/(x-2) lnx
so f(2) = -ln 2
Thanks Steve ur a life saver!!
1. To determine which function is not continuous everywhere, we need to check if there are any points where the function is discontinuous. A function is continuous at a particular point if the limit from the left is equal to the limit from the right, and if the function value at that point matches the limit.
A. y = (4x)/(x+1)^2:
This function is a rational function, and it is continuous everywhere except at the points where the denominator becomes zero. In this case, the denominator (x+1)^2 becomes zero when x = -1. So, function A is continuous everywhere except at x = -1.
B. y = |x|:
The absolute value function is continuous everywhere, including when x = 0.
C. y = (x)/(x^2+1):
This function is also a rational function, and it is continuous everywhere except at the points where the denominator becomes zero. In this case, the denominator (x^2+1) becomes zero when x = ±i (where i is the imaginary unit). Since these values are not on the real number line, function C is continuous everywhere.
D. y = √(x^2+1):
This is a square root function, and it is continuous everywhere because the square root of a positive number is defined for all real numbers. So, function D is continuous everywhere.
E. y = x^(2/3):
This is a power function, and it is continuous everywhere. Power functions are continuous for all real numbers, including negative values and zero.
Based on the above analysis, the function that is not continuous everywhere is A. y = (4x)/(x+1)^2, because it has a discontinuity at x = -1.
2. The domain of a function represents the set of all possible input values (x) for which the function is defined. To find the domain of f(x) = √((x+6))sin(x), we need to consider two main factors:
a) For the square root function (√), the argument within the square root sign must be non-negative. This means that (x+6) ≥ 0, leading to x ≥ -6.
b) For the sine function (sin), there are no restrictions on the domain. It is defined for all real numbers.
Combining these conditions, we find that the domain of f(x) is all real numbers greater than or equal to -6, denoted as (-6, ∞).
3. To find the value of 'a' so that the function h(x) is continuous at x = 0, we need to check if the function approaches the same value from both sides of x = 0.
Given h(x) = (sin(2x))/x for x ≠ 0, and h(x) = a for x = 0.
Taking the limit as x approaches 0 from the left (x → 0^-):
lim(x → 0^-) [(sin(2x))/x] = 1
As the limit exists and is equal to 1, we can set 'a' to be equal to 1.
Therefore, the value of 'a' that makes h(x) continuous at x = 0 is b. a = 1.
4. To find the value of f(2), we need to substitute x = 2 into the given function f(x) = (ln(x^2) - xln(x))/(x - 2) when x ≠ 2.
When x = 2, the expression (x - 2) in the denominator becomes 0, resulting in an undefined value.
However, if we take the limit of the function as x approaches 2, we can determine the value of f(2) using the concept of continuity.
lim(x → 2) [(ln(x^2) - xln(x))/(x - 2)]
= lim(x → 2) [(2ln(x) - xln(x))/(x - 2)]
= lim(x → 2) [(2ln(x) - xln(x))/(x - 2)] [multiply by (x + 2)/(x + 2)]
= lim(x → 2) [(2ln(x) - xln(x))(x + 2)/(x - 2)(x + 2)]
= lim(x → 2) [(2ln(x)(x + 2) - xln(x)(x + 2))/(x^2 - 4)]
= 2ln(2)(2 + 2)/(4 - 4) [substituting x = 2]
= 4ln(2)/0
= ∞
Therefore, f(2) is undefined or approaches infinity.