Mar 14, 2018 · x^2-2x+1 This is a fixed identity, these types of expressions, knowing the theory behind them, can be multiplied easily without any calculations. For example, let's say you have the expression (x+y)^2 It becomes: x^2+2xy+y^2 If you have (x-y)^2 It becomes: x^2-2xy+y^2 You can see the pattern, you can learn it by heart, it is always the same in these cases. But to …

Aug 3, 2015 · Domain: (-oo, ln(2)) uu (ln(2), +oo) Range: (-oo, 0) uu (1/2, +oo) Right from the start, you can say that one value of x will be excluded from the domain of the function because of the restriction to the denominator of the function. More epecifically, the expression (2-e^x) must not be equal to zero. The value of x for which condition is not satisfied will be 2-e^x = 0 e^x = 2 …

Feb 17, 2016 · Vertical asymptote at x=-1. No horizontal asymptotes. Vertical asymptote at places which make the denominator zero, ie when x=-1.

Dec 12, 2016 · x+1/(2(x-1)) +1/(2(x+1)) How do you write the partial fraction decomposition of the rational expression # (3x^2 + 12x -...

Nov 22, 2016 · d/dxarctan(x^2 +1) = (1/(1+(x^2+1)^2))*2x Consider the derivative of arctan(u) d/dx arctan(u) = (1/(1+u^2)) * d/dx (u) In your question: u = (x^2+1)

Mar 13, 2018 · color(blue)(y=x+1) To find the equation of the tangent line at x=-1 we must first find the gradient of the line at the given point. If we differentiate f(x) this will give us a gradient function and allow us to find the gradient at any point on f(x) Differentiating f(x)=x^2+3x+2 using the Power Rule: dy/dx(ax^n)=anx^(n-1) Note that dy/dx is distributive over the sum: i.e …

Sep 15, 2014 · For this problem we have 3 options. 1) We can use the completing the square the method to put the equations in Standard Vertex Form, (x-h)^2-k=0 -> vertex (h,-k) 2) We can use the expression (-b/(2a),f(-b/(2a))) where a and b are the coefficients found when the equation is in standard form, ax^2+bx+c=0 3) We could just graph it. Let do options 1 & 2. Completing the …

Range is [ 0, pi/2 ]. Domain for x is [ 0, 2 ]. Also introduced is the inverse operator (cos)^(-1), on par with f^(-1). Over centuries, we have been told that the range of cos^(-1)x or, for that matter, arccos x is [ 0, pi ]. In the f^(-1) sense, I like to use (cos)^(-1) to state that the range of (cos)^(-1) x is ( - oo, oo ). Here, the conventional range of y = arccos( ( x - 1 )^2) is [ 0, pi ...

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Aug 11, 2015 · {(x = 6), (x=2) :} Start by writing the conditions that a value of x must satisfy in order to be a valid solution to this equation. You have to do that because you're dealing with radical terms, which means that the expressions under the radical signs must be positive for real numbers. 4x+1>=0 implies x >=-1/4 x-2>=0 implies x >= 2 Combine these two conditions to …