y=x+3/[(x-4)(x+3)] Definition: a critical number of a function f is a number c in the domain of f such that f'( c) = 0 or f'(c) doesn't exist. Since we have a rational function, the domain is all real numbers except the numbers that make the denominator zero. Therefore, x cannot be -3 and 4. There is a hole on the graph of the function at x = -3, and x = 4 is a vertical asymptote. Simplify first, then take the derivative of y. y = (x + 3)/[(x - 4)(x + 3)] = 1/(x - 4) y' = [1/(x - 4)]' use the quotient rule y' = [(x - 4)(0) - 1(1)]/(x - 4)^2 = -1/(x - 4)^2. Since the numerator is a constant, then y' is never zero, so there is not a critical point.
y=2x+4 --> x=2y+4 ==> y=(x-4)/2
Start with y = |x|, then y = 4|x|, and then y = -4|x|.
Get in function form. - 3X - Y = 4 - 1(- Y = 3X + 4) Y = - 3X - 4 ----------------------------solve for X and Y by the 0 out method - 3X - 4 = 0 - 3X = 4 X = - 4/3 --------------- Y = - 3(0) - 4 Y = - 4 -------------- Draw a line linking those points.
Let f(x) = y y = 1 + (4/x) Now replace y with x and x with y and find equation for y x = 1 + (4/y) (x-1) = (4/y) y = 4/(x-1) This g(x), the inverse of f(x) g(x)= 4/(x-1) The domain will be all real numbers except when (x-1)=0 or x=1 So Domain = (-ā,1),(1,+ā) And Range = (-ā,0),(0,+ā) f(g(x)) = f(4/(x-1)) = 1 + 4/(4/(x-1)) = 1+(x-1) = x g(f(x)) = g(1+(4/x)) = 4/((1+(4/x))-1) = 4/(4/x) = x So we get f(g(x)) = g(f(x)) Notice the error in copying the next part of your question It should be g'(f(x)) = 1/(f'(g(x))) g'(f(x)) = d/dx (g(f(x))) = d/dx (x) = 1 f'(g(x)) = d/dx (f(g(x))) = d/dx (x) = 1 1/[f'(g(x))] = 1/1 = 1 g'(f(x)) = 1/f'(g(x)) ( Notice the error in copying your question)
A function is an equation that gives a unique answer. A relation does not. Example: y = 3x + 1 is a function. If I give you x, you can determine y. And that y is unique to that x. So if x = 1, you know y = 4. No other of x gives y = 4 as an answer. So y = 3x + 1 is a function. Example: y = 4x2. So if I give you x = 1, y = 4. But y = 4 if I also give you x = -1. So y = 4x2 is not a function, it is a relation.
1
Yes, y=x-4 is linear, since the highest power on x is 1 (y = x1-4)
-6
Y = x2
The inverse of a function (G(x)) can be found by switching the roles of (x) and (y) and solving for (y). Given the function (G(x) = -\frac{4}{3}x + 2), let's find its inverse: Step 1: Replace (G(x)) with (y): [y = -\frac{4}{3}x + 2] Step 2: Swap (x) and (y): [x = -\frac{4}{3}y + 2] Step 3: Solve for (y): [x - 2 = -\frac{4}{3}y] [-\frac{3}{4}(x - 2) = y] So, the inverse function (G^{-1}(x)) is: [G^{-1}(x) = -\frac{3}{4}(x - 2)]
X - Y^2 = 1 - Y^2 = - X + 1 Y^2 = X - 1 Y = (+/-) sqrt(X - 1) now, X is represented as a function of Y. Function values are generally Y values.
The function y = x is the graph that passes from the points (-1, -1), (0, 0), and (1, 1) The function y = 4x is the graph that passes form the points (-1, -4), (0, 0), and (1, 4) Sketch these graphs in a same x and y coordinate system, and you can see both of them
y = x(x + 1)
Yes it does, Remember Y values are generally function values. So, putting a value into this function, substitution a integer for X, fives you the Y value. Y = X + 4 ( make X 2 ) Y = (2) + 4 Y = So, when X = 2, Y = 6. The function.
To find the inverse of a function, you have to switch y for x and then solve for y again. These are the steps: y = 5x - 20 x = 5y - 20 5y = x + 20 y = 0.2x + 4 or f(x)^-1 = 0.2x + 4
Consider it as a graph with points (x, f(x)): you have two points (0, 1) and (1, 4) joined by a line (as it is a linear function). The slope m = change in y/change in x = (4-1)/(1-0) = 3 Using y - Yo = m(y - Xo) y - 1 = 3(x - 0) → y = 3x + 1 → f(x) = 3x + 1