(1) State the conditions for a singled-valued function of x to be continuous at x = a
(2) find the points of discontinuity of the functions
(i) x^3 + 4x + 6 / x^2 - 6x + 8
(ii) sec x
(3) Differentiate from the first princiaples (i) x^3 + 2x - 1
(1)
If we take the meaning of a "a singled-valued function" in the normal sense of a function (that passes the vertical line test), the conditions for continuity of f(x) at point a are:
1. f(a) exists.
2. Lim x→a+ exists and Lim x→a- exists.
3. Lim x→a+ = Lim x→a-.
(2)
If f(x) is a quotient of two polynomials, f(x) is continuous within its own domain, which excludes the vertical asymptotes. The asymptotes occur where the denominator becomes zero.
Since sec(x) = 1/cos(x), the same idea applies.
(3)
In case it is needed,
(x+h)³=x³+3x²h+3xh²+h³
solve the question for me
(1) To determine the conditions for a single-valued function of x to be continuous at x = a, we need to ensure that three conditions are met:
- The function is defined at x = a.
- The limit of the function as x approaches a exists.
- The value of the function at x = a equals the limit.
In mathematical terms, for a single-valued function f(x) to be continuous at x = a, the following conditions must hold:
1. f(a) must be defined.
2. lim(x → a) f(x) must exist.
3. f(a) = lim(x → a) f(x).
(2) To find the points of discontinuity of the given functions, we need to identify the values of x for which the functions fail to meet the continuity conditions.
(i) For the function f(x) = (x^3 + 4x + 6) / (x^2 - 6x + 8), we need to identify the values of x where the denominator becomes zero. Since the denominator is a quadratic expression, we solve the equation x^2 - 6x + 8 = 0 to find the roots.
Factoring or using the quadratic formula, we have (x - 2)(x - 4) = 0, which gives x = 2 and x = 4 as the roots.
Therefore, the points of discontinuity for this function are x = 2 and x = 4.
(ii) For the function f(x) = sec(x), we need to consider the values of x for which the function is undefined. Since the secant function is the reciprocal of the cosine function, it is undefined when the cosine function is zero.
The cosine function is equal to zero at x = π/2, x = 3π/2, x = 5π/2, etc. These are the points of discontinuity for the function sec(x).
(3) To differentiate the function f(x) = x^3 + 2x - 1 using the first principles (also known as the definition of the derivative), we follow these steps:
1. Start with the definition of the derivative:
f'(a) = lim(h → 0) [f(a + h) - f(a)] / h.
2. Substitute the function f(x) = x^3 + 2x - 1 into the definition of the derivative:
f'(a) = lim(h → 0) [(a + h)^3 + 2(a + h) - 1 - (a^3 + 2a - 1)] / h.
3. Simplify the equation:
f'(a) = lim(h → 0) [a^3 + 3a^2h + 3ah^2 + h^3 + 2a + 2h - 1 - a^3 - 2a + 1] / h.
4. Cancel out common terms and simplify:
f'(a) = lim(h → 0) [3a^2h + 3ah^2 + h^3 + 2h] / h.
f'(a) = lim(h → 0) [3a^2 + 3ah + h^2 + 2].
5. Take the limit as h approaches 0:
f'(a) = 3a^2 + 2.
Therefore, the derivative of f(x) = x^3 + 2x - 1 from the first principles is f'(x) = 3x^2 + 2.