z: alternate interior angles
c: consecutive interior angles
f: corresponding angles
x: vertically opposite angles
ex+f = c -dx ex+dx = c -f x(e+d) = c -f x = c -f/(e+d)
While no set of rules can handle differentiating every expression, the following should help. For all of the following, assume c and n are constants, f(x) and g(x) are functions of x, and f'(x) and g'(x) mean the derivative of f and g respectively. Constant derivative rule:d/dx(c)=0 Constant multiple rule:d/dx(c*f(x))=c*f'(x) Sum and Difference Rule:d/dx(f(x)±g(x))=f'(x)±g'(x) Power rule:d/dx(xn)=n*xn-1 Product rule:d/dx(f(x)*g(x))=f'(x)*g(x) + g'(x)*f(x) Quotient rule:d/dx(f(x)/g(x))=(f'(x)*g(x)-g'(x)*f(x))/f(x)² Chain rule:d/dx(f(g(x))= f'(g(x))*g'(x)
∫ f'(x)/[f(x)√(f(x)2 - a2)] dx = (1/a)arcses(f(x)/a) + C C is the constant of integration.
f(x) = 2 * 2 - x + 9 f(-4) = 2 * 2 -(-4) + 9 f(-4) = 4 + 4 + 9 = 17
Simply integrate all the pieces apart, en add them up. This is allowed, because int_a^c f(x)dx = int_a^b f(x)dx + int_b^c f(x)dx for all a,b,c in dom(f).
Let f be a function with domain D in R, the real numbers, and D is an open set in R. Then the derivative of f at the point c is defined as: f'(c) =lim as x-> c of the difference quotient [f(x)-f(c)]/[x-c] If that limit exits, the function is called differentiable at c. If f is differentiable at every point in D then f is called differentiable in D.
A function f is continuous at c if:f(c) is defined.lim "as x approaches c" f(x) exists.lim "as x approaches c" f(x) = f(c).
∫ f(x)/[(f(x) + b)(f(x) + c)] dx = [b/(b - c)] ∫ 1/(f(x) + b) dx - [c/(b - c)] ∫ 1/(f(x) + c) dx b ≠c
A Maclaurin series is centered about zero, while a Taylor series is centered about any point c. M(x) = [f(0)/0!] + [f'(0)/1!]x +[f''(0)/2!](x^2) + [f'''(0)/3!](x^3) + . . . for f(x). T(x) = [f(c)/0!] + [f'(c)/1!](x-c) +[f''(c)/2!]((x-c)^2) + [f'''(c)/3!]((x-c)^3) + . . . for f(x).
Yes, vertical angles are formed by an x. Opposite angles are the angles opposite on the x. For instance...AB X CDAnd and d are opposite angles, so are B and C. This whole figure is called a vertical angle.
ex+f = c -dx ex+dx = c -f x(e+d) = c -f x = c -f/(e+d)
Critical point (in one variable): A point on the graph y = f(x) at which f is differentiable and f'(x) = 0. The term is also used for the number c such that f'(c) = 0. The corresponding value f(c) is a critical value. A critical point c can be classified depending upon the behavior of fin the neighborhood of c, as one of following:1. a local minimum, if f'(x) > 0 to the left of c and f'(x)< 0 to the right of c.2. a local maximum, if f'(x) < 0 to the left of c and f'(x)> 0 to the right of c.3. neither local maximum nor local minimum:a) if f'(x) has the same sign to the left and to the right of c, in which case c is a horizontal point of inflection.b) if there is an interval at every point of which f'(x) = 0 and c is an endpoint or interior point of this interval.(This is called the first derivative test).The second derivative test (is also a test for maximum and minimum values. It is a consequence of the concavity test):Suppose f is continuous near c.1. If f'(c) = 0 and f''(c) > 0, then f has a local minimum at c.2. If f'(c) = 0 and f''(c) < 0, then f has a local maximum at c.Example: f(x) = x^3 - 12x + 1(a) Find the intervals on which f is increasing or decreasing.(b) Find the local maximum and minimum values of f.(c) Find the intervals of concavity and the inflection points.Solution:(a) f(x) = x^3 - 12x + 1f'(x) = 3x^2 - 12f'(x) = 3(x +2)(x - 2)Interval: x < -2; -2 x < 2; x > 2x + 2: - ; +; +x - 2: - ; - ; +f'(x): + ; - ; +f: increasing on (-∞, -2); decreasing on (-2, 2); increasing on (2, ∞) So f is increasing on (-∞, -2) and (2, ∞) and f is decreasing on (-2, 2).(b) f changes from increasing to decreasing at x = -2 and from decreasing to increasing at x = 2. Thus f(-2) = 17 is a local maximum value and f(2) = -15 is a local minimum value.(c) f''(x) = 6xf''(x) > 0 ↔ x > 0 and f''(x) < 0 ↔ x < 0. Thus f is concave upward on (0, ∞) and concave downward on (-∞, 0). There is an inflection point where the concavity changes, at (0, f(0)) = (0, 1).
Using calculus to see if the function f(x) is continuous at a point (point c) involves three steps. These three conditions must be met: 1. f(c) exists, is defined 2. lim f(x) exists x-->c 3. f(c)= lim f(x) x-->c
�F = �C x 9/5 + 32 or, in reverse, �C = (�F - 32) X 5/9
-17.22 C °C x 9/5 + 32 = °F (°F - 32) x 5/9 = °C
Undefined. Proof, let a function f have a vertical line on x = c. (Notice: By definition of functions, it is not even a function, which means that we do not even need to discuss differentiability. We assume it is a function) Now suppose f'(c) exist, then f'(c) = lim x --> c (f(x) - f(c))/(x - c), the limit exist but since it's a straight line, assume non trivial (a point), we have automatically x = c. But since it's non-trivial, hence f(x) != f(c), let f(x) - f(c) = r for some real number r != 0. we get f'(c) = lim x --> c (f(x) - f(c)) / (x - c) = r/0 which is undefined. Contradiction!. Hence f'(c) doesn't exist. Note: If you see a straight line some where on a "function" and they ask for derivative, write:"This is not a function, what kind of question is this! Go back to Calculus class!" If you are discussing a function with a vertical slope, e.g. let f(x) = cubeRoot(x), then it's a different proof.
∫ f'(x)/f(x) dx = ln(f(x)) + C C is the constant of integration.