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Inverse function 2x plus 4?
y=2x+4 --> x=2y+4 ==> y=(x-4)/2
What is the equation for the absolute value function stretched vertically by a value of 4 and reflected across the x axis?
Start with y = |x|, then y = 4|x|, and then y = -4|x|.
How do you graph -3x-y equals 4?
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.
Determine the inverse g(x) of the function f(x) = (1+(4/x)), stating its domain and range. Verify that f(g(x)) = g(f(x))=x and that g′(f(x)) = (1/(f′(x))?
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 = (-∞,1),(1,+∞) And Range = (-∞,0),(0,+∞) 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)
Is the point of discontinuity considered as critical point?
Well, honey, a point of discontinuity is not the same as a critical point in calculus. A critical point is where the derivative is either zero or undefined, while a point of discontinuity is where a function is not continuous. So, in short, they may both be important in their own ways, but they're not the same thing.
