the linearization of f(x) = x^4 + 3x^2 with a = -1 is L(x) = 4 - 10(x+1)
the linearization formula is L(x) = f(a) + f'(a)*(x-a)
when, a = -1
f(a) = (-1)^4 + 3(-1)^2 = 4
f'(x) = 4*x^3 + 6*x
f'(a) = 4(-1)^3 + 6(-1) = -10
so, L(x) = (4) + (-10)*(x-(-1)) = 4 - 10(x+1)
The square root operation is not a function because for each value of y there can be 2 values of x - the principal square root and its negative. This can only be rectified by limiting the range of the opearation to the principal or positive square root. Furthermore, it also depends on the domain of the function. If y<4 then the square root is not defined within Real numbers. So, for y ≥ 4, x = +sqrt(y-4) is a function.
5
The square root of two times the square root of two equals two
Find the range of a function by substituting the highest domain possible and the lowest domain possible into the function. There, you will find the highest and lowest range. Then, you should check all the possible cases in the function where a number could be divided by 0 or a negative number could be square rooted. Remove these numbers from the range. A good way to check to see if you have the correct range is to graph the function (within the domain, of course).
when x is a negative number --- is a wrong answer since square root of a negative number is not defined. So x has to be zero or a positive number. The correct answer is that when x lies between 0 and 1 (with both limits excluded), its square root is greater than the number itself. Of course at both limits, the square root (assuming the positive square root - since a square root of a number can be positive or negative, both with the same absolute value) is the same as the number.
i = the square root of negative one
-2
x = 4
Square root of s, or negative square root of s.
The square root operation is not a function because for each value of y there can be 2 values of x - the principal square root and its negative. This can only be rectified by limiting the range of the opearation to the principal or positive square root. Furthermore, it also depends on the domain of the function. If y<4 then the square root is not defined within Real numbers. So, for y ≥ 4, x = +sqrt(y-4) is a function.
Negative 1.047197551 etc, etc.
Every positive real number has two square roots: one negative and one positive. As a result, the square root mapping is one-to-many and so is not a mathematical function. One way to make it a function is to restrict the range to non-negative real numbers. These are the non-negative square roots.
No.
They can, but they are called imaginary numbers. This is because a "square" of something is the number that, when multiplied by itself, will equal the first number. A negative multiplied by a negative equals a positive, and positives can't turn into negatives on their own, so there are no square roots of negative numbers.
square root of x/pi
A positive number has two square roots, its principal (positive) root and its negative root. This is because a negative number multiplied by a negative number equals a positive number. In this instance, the square roots of 256 are 16 and -16.
Yes, a negative sign has a square root.This is done through using the imaginary unit defined as: i = .for example the square root of -36 is square root of (-1) multiplied by square root of 36. Accordingly it equals 6i