Answer to Question #185118 in Calculus for Talha

Question #185118

Find equations of all lines having slope -1 that are tangent to the curve 𝑦=1/π‘₯βˆ’1



1
Expert's answer
2021-04-27T01:24:26-0400

Given the equation y=1xβˆ’1,y=\frac{1}{x-1},


m=dydx=βˆ’1(xβˆ’1)2=βˆ’1(given)m=\dfrac{dy}{dx}=-\frac{1}{(x-1)^2}=-1 \qquad \text{(given)}

β€…β€ŠβŸΉβ€…β€Š1(xβˆ’1)2=11=(xβˆ’1)21=x2βˆ’2x+11βˆ’1=x2βˆ’2x0=x(xβˆ’2)β€…β€ŠβŸΉβ€…β€Šx=0 and x=2\implies \frac{1}{(x-1)^2} = 1\\ 1= (x-1)^2\\ 1= x^2-2x+1\\ 1-1=x^2-2x\\ 0=x(x-2)\\ \implies x=0 \text{ and } x=2

For x=0,x=0,

y=10βˆ’1=1βˆ’1=βˆ’1∴the tangent line touches the curve at the point (0,-1)y=\frac{1}{0-1}=\frac{1}{-1}=-1\\ \therefore \text{the tangent line touches the curve at the point (0,-1)}


The equation of the line is:

yβˆ’y1=m(xβˆ’x1)yβˆ’(βˆ’1)=βˆ’1(xβˆ’0)y+1=βˆ’xβ€…β€ŠβŸΉβ€…β€Šx+y=βˆ’1β‹―(i)y-y_1=m(x-x_1)\\ y-(-1)=-1(x-0)\\ y+1=-x\\ \implies x+y=-1 \qquad \cdots (i)

at x=2,x=2,

y=12βˆ’1=11=1∴the tangent line touches the curve at the point (2,1)y=\frac{1}{2-1}=\frac{1}{1}=1\\ \therefore \text{the tangent line touches the curve at the point (2,1)}

yβˆ’y1=m(xβˆ’x1)yβˆ’(βˆ’1)=βˆ’1(xβˆ’2)yβˆ’1=βˆ’x+2β€…β€ŠβŸΉβ€…β€Šx+y=3β‹―(ii)y-y_1=m(x-x_1)\\ y-(-1)=-1(x-2)\\ y-1=-x+2\\ \implies x+y=3 \qquad \cdots (ii)

Thus, eqn (i) and (ii) are the required equations of the line that are tangential to the curve.


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