Equation of the tangent to a function
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Batione1995
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Equation of the tangent to a function
Write the equation of tangents for the graph of the function \(f(x)={x−1\over x+1}\) and parallel to the line with the equation \(y=2x+1\).
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janusz55
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Re: Equation of the tangent to a function
\( f(x) = \frac{x-1}{x+1}. \)
\( D= \{x: x\in\rr \wedge x\neq -1\} = \rr \setminus \{-1\}.\)
Equation of tangent
\( y = f'(p)(x-p) + f(p) \)
The tangent is supposed to be parallel to the given straight line \( y = 2x +1.\) Their coefficients of direction must be equal \( 2.\)
\( f'(p) = \frac{1\cdot (p+1)- (p-1)\cdot 1}{(p+1)^2} = \frac{2}{(p-1)^2} = 2.\)
\( (p-1)^2 = 1, \ \ p-1 = -1 \ \ \vee p-1 = 1, \ \ p_{1} = 0, \ \ p_{2} = 2.\)
Equations of tangent
\( y = 2(x -0) + f(0) = 2x -1, \)
\( y = 2(x -2) + f(2) = 2x - 4 +\frac{1}{3} = 2x -\frac{11}{3}.\)
\( D= \{x: x\in\rr \wedge x\neq -1\} = \rr \setminus \{-1\}.\)
Equation of tangent
\( y = f'(p)(x-p) + f(p) \)
The tangent is supposed to be parallel to the given straight line \( y = 2x +1.\) Their coefficients of direction must be equal \( 2.\)
\( f'(p) = \frac{1\cdot (p+1)- (p-1)\cdot 1}{(p+1)^2} = \frac{2}{(p-1)^2} = 2.\)
\( (p-1)^2 = 1, \ \ p-1 = -1 \ \ \vee p-1 = 1, \ \ p_{1} = 0, \ \ p_{2} = 2.\)
Equations of tangent
\( y = 2(x -0) + f(0) = 2x -1, \)
\( y = 2(x -2) + f(2) = 2x - 4 +\frac{1}{3} = 2x -\frac{11}{3}.\)