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Question
y=\frac{x^{2}+x+1}{x}
Function
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Evaluate the derivative
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Find the domain
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\text{Find the }x\text{-intercept/zero}
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Find the y-intercept
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Find the critical numbers
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Find the local extrema
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Find the increasing or decreasing interval
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Find the range
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Find the vertical asymptotes
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Find the horizontal asymptotes
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Find the oblique asymptotes
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Determine if even, odd or neither
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Find the stationary points
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Find the inflection points
More methods
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y^{\prime}=\frac{x^{2}-1}{x^{2}}
Evaluate
y=\frac{x^{2}+x+1}{x}
Take the derivative of both sides
y^{\prime}=\frac{d}{dx}\left(\frac{x^{2}+x+1}{x}\right)
\text{Use differentiation rule }\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{\frac{d}{dx}(f(x))\times g(x)-f(x)\times \frac{d}{dx}(g\left(x)\right)}{\left(g(x)\right)^2}
y^{\prime}=\frac{\frac{d}{dx}\left(x^{2}+x+1\right)\times x-\left(x^{2}+x+1\right)\times \frac{d}{dx}\left(x\right)}{x^{2}}
Calculate
More Steps
Evaluate
\frac{d}{dx}\left(x^{2}+x+1\right)
\text{Use differentiation rule }\frac{d}{dx}\left(f\left(x\right)\pm g\left(x\right)\right)=\frac{d}{dx}\left(f\left(x\right)\right)\pm \frac{d}{dx}(g(x))
\frac{d}{dx}\left(x^{2}\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(1\right)
\text{Use }\frac{d}{dx} x^{n}=n x^{n-1}\text{ to find derivative}
2x+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(1\right)
\text{Use }\frac{d}{dx} x^{n}=n x^{n-1}\text{ to find derivative}
2x+1+\frac{d}{dx}\left(1\right)
\text{Use }\frac{d}{dx}(c)=0\text{ to find derivative}
2x+1+0
Removing 0 doesn't change the value,so remove it from the expression
2x+1
y^{\prime}=\frac{\left(2x+1\right)x-\left(x^{2}+x+1\right)\times \frac{d}{dx}\left(x\right)}{x^{2}}
\text{Use }\frac{d}{dx} x^{n}=n x^{n-1}\text{ to find derivative}
y^{\prime}=\frac{\left(2x+1\right)x-\left(x^{2}+x+1\right)\times 1}{x^{2}}
Multiply the terms
y^{\prime}=\frac{x\left(2x+1\right)-\left(x^{2}+x+1\right)\times 1}{x^{2}}
Any expression multiplied by 1 remains the same
y^{\prime}=\frac{x\left(2x+1\right)-\left(x^{2}+x+1\right)}{x^{2}}
Solution
More Steps
Evaluate
x\left(2x+1\right)-\left(x^{2}+x+1\right)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
x\left(2x+1\right)-x^{2}-x-1
Expand the expression
2x^{2}+x-x^{2}-x-1
Subtract the terms
x^{2}+x-x-1
The sum of two opposites equals 0
x^{2}+0-1
Remove 0
x^{2}-1
y^{\prime}=\frac{x^{2}-1}{x^{2}}
Show Solutions
Testing for symmetry
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Testing for symmetry about the origin
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Testing for symmetry about the x-axis
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Testing for symmetry about the y-axis
\textrm{Not symmetry with respect to the origin}
Evaluate
y=\frac{x^{2}+x+1}{x}
\text{To test if the graph of }y=\frac{x^{2}+x+1}{x}\text{ is symmetry with respect to the origin,substitute -x for x and -y for y}
-y=\frac{\left(-x\right)^{2}-x+1}{-x}
Simplify
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Evaluate
\frac{\left(-x\right)^{2}-x+1}{-x}
Rewrite the expression
\frac{x^{2}-x+1}{-x}
\text{Use }\frac{-a}{b}=\frac{a}{-b}=-\frac{a}{b}\text{ to rewrite the fraction}
-\frac{x^{2}-x+1}{x}
-y=-\frac{x^{2}-x+1}{x}
Change the signs both sides
y=\frac{x^{2}-x+1}{x}
Solution
\textrm{Not symmetry with respect to the origin}
Show Solutions
Solve the equation
\begin{align}&x=\frac{\sqrt{-3-2y+y^{2}}-1+y}{2}\\&x=\frac{-\sqrt{-3-2y+y^{2}}-1+y}{2}\end{align}
Evaluate
y=\frac{x^{2}+x+1}{x}
Swap the sides of the equation
\frac{x^{2}+x+1}{x}=y
Cross multiply
x^{2}+x+1=xy
Simplify the equation
x^{2}+x+1=yx
Move the expression to the left side
x^{2}+x+1-yx=0
Collect like terms by calculating the sum or difference of their coefficients
x^{2}+\left(1-y\right)x+1=0
Move the constant to the right side
x^{2}+\left(1-y\right)x=0-1
Add the terms
x^{2}+\left(1-y\right)x=-1
Add the same value to both sides
x^{2}+\left(1-y\right)x+\frac{1-2y+y^{2}}{4}=-1+\frac{1-2y+y^{2}}{4}
Evaluate
x^{2}+\left(1-y\right)x+\frac{1-2y+y^{2}}{4}=\frac{-3-2y+y^{2}}{4}
Evaluate
\left(x+\frac{1-y}{2}\right)^{2}=\frac{-3-2y+y^{2}}{4}
Take the root of both sides of the equation and remember to use both positive and negative roots
x+\frac{1-y}{2}=\pm \sqrt{\frac{-3-2y+y^{2}}{4}}
Simplify the expression
More Steps
Evaluate
\sqrt{\frac{-3-2y+y^{2}}{4}}
To take a root of a fraction,take the root of the numerator and denominator separately
\frac{\sqrt{-3-2y+y^{2}}}{\sqrt{4}}
Simplify the radical expression
More Steps
Evaluate
\sqrt{4}
\text{Write the number in exponential form with the base of }2
\sqrt{2^{2}}
\text{Reduce the index of the radical and exponent with }2
2
\frac{\sqrt{-3-2y+y^{2}}}{2}
x+\frac{1-y}{2}=\pm \frac{\sqrt{-3-2y+y^{2}}}{2}
\text{Separate the equation into }2\text{ possible cases}
\begin{align}&x+\frac{1-y}{2}=\frac{\sqrt{-3-2y+y^{2}}}{2}\\&x+\frac{1-y}{2}=-\frac{\sqrt{-3-2y+y^{2}}}{2}\end{align}
Calculate
More Steps
Evaluate
x+\frac{1-y}{2}=\frac{\sqrt{-3-2y+y^{2}}}{2}
Move the expression to the right-hand side and change its sign
x=\frac{\sqrt{-3-2y+y^{2}}}{2}-\frac{1-y}{2}
Subtract the terms
More Steps
Evaluate
\frac{\sqrt{-3-2y+y^{2}}}{2}-\frac{1-y}{2}
Write all numerators above the common denominator
\frac{\sqrt{-3-2y+y^{2}}-\left(1-y\right)}{2}
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
\frac{\sqrt{-3-2y+y^{2}}-1+y}{2}
x=\frac{\sqrt{-3-2y+y^{2}}-1+y}{2}
\begin{align}&x=\frac{\sqrt{-3-2y+y^{2}}-1+y}{2}\\&x+\frac{1-y}{2}=-\frac{\sqrt{-3-2y+y^{2}}}{2}\end{align}
Solution
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Evaluate
x+\frac{1-y}{2}=-\frac{\sqrt{-3-2y+y^{2}}}{2}
Move the expression to the right-hand side and change its sign
x=-\frac{\sqrt{-3-2y+y^{2}}}{2}-\frac{1-y}{2}
Subtract the terms
More Steps
Evaluate
-\frac{\sqrt{-3-2y+y^{2}}}{2}-\frac{1-y}{2}
Write all numerators above the common denominator
\frac{-\sqrt{-3-2y+y^{2}}-\left(1-y\right)}{2}
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
\frac{-\sqrt{-3-2y+y^{2}}-1+y}{2}
x=\frac{-\sqrt{-3-2y+y^{2}}-1+y}{2}
\begin{align}&x=\frac{\sqrt{-3-2y+y^{2}}-1+y}{2}\\&x=\frac{-\sqrt{-3-2y+y^{2}}-1+y}{2}\end{align}
Show Solutions
Rewrite the equation
\begin{align}&r=\frac{\cos\left(\theta \right)+\sqrt{-3\cos^{2}\left(\theta \right)+2\sin\left(2\theta \right)}}{\sin\left(2\theta \right)-2\cos^{2}\left(\theta \right)}\\&r=\frac{\cos\left(\theta \right)-\sqrt{-3\cos^{2}\left(\theta \right)+2\sin\left(2\theta \right)}}{\sin\left(2\theta \right)-2\cos^{2}\left(\theta \right)}\end{align}
Evaluate
y=\frac{x^{2}+x+1}{x}
Multiply both sides of the equation by LCD
yx=\frac{x^{2}+x+1}{x}\times x
Simplify the equation
yx=x^{2}+x+1
Move the expression to the left side
yx-x^{2}-x=1
\text{To convert the equation to polar coordinates,substitute }r\cos\left(\theta \right)\text{ for }x\text{ and }r\sin\left(\theta \right)\text{ for }y
\sin\left(\theta \right)\times r\cos\left(\theta \right)\times r-\left(\cos\left(\theta \right)\times r\right)^{2}-\cos\left(\theta \right)\times r=1
Factor the expression
\left(\sin\left(\theta \right)\cos\left(\theta \right)-\cos^{2}\left(\theta \right)\right)r^{2}-\cos\left(\theta \right)\times r=1
Simplify the expression
\left(\frac{1}{2}\sin\left(2\theta \right)-\cos^{2}\left(\theta \right)\right)r^{2}-\cos\left(\theta \right)\times r=1
Subtract the terms
\left(\frac{1}{2}\sin\left(2\theta \right)-\cos^{2}\left(\theta \right)\right)r^{2}-\cos\left(\theta \right)\times r-1=1-1
Evaluate
\left(\frac{1}{2}\sin\left(2\theta \right)-\cos^{2}\left(\theta \right)\right)r^{2}-\cos\left(\theta \right)\times r-1=0
Solve using the quadratic formula
r=\frac{\cos\left(\theta \right)\pm \sqrt{\left(-\cos\left(\theta \right)\right)^{2}-4\left(\frac{1}{2}\sin\left(2\theta \right)-\cos^{2}\left(\theta \right)\right)\left(-1\right)}}{\sin\left(2\theta \right)-2\cos^{2}\left(\theta \right)}
Simplify
r=\frac{\cos\left(\theta \right)\pm \sqrt{-3\cos^{2}\left(\theta \right)+2\sin\left(2\theta \right)}}{\sin\left(2\theta \right)-2\cos^{2}\left(\theta \right)}
Solution
\begin{align}&r=\frac{\cos\left(\theta \right)+\sqrt{-3\cos^{2}\left(\theta \right)+2\sin\left(2\theta \right)}}{\sin\left(2\theta \right)-2\cos^{2}\left(\theta \right)}\\&r=\frac{\cos\left(\theta \right)-\sqrt{-3\cos^{2}\left(\theta \right)+2\sin\left(2\theta \right)}}{\sin\left(2\theta \right)-2\cos^{2}\left(\theta \right)}\end{align}
Show Solutions
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