Find equations of all lines having slope -1 that are tangent to the curve y=1 /(x-1).

Name: Problem 27, Chapter 3: University Calculus: Early Transcendentals
Uploaded: 2020-12-09T05:34:00-08:00
Duration: 6 min 14 s
Channel: Bobby Barnes
Description: Find equations of all lines having slope -1 that are tangent to the curve y=1 /(x-1).

Find equations of all lines having slope -1 that are tangent...

Key Concepts

Tangent Line

A tangent line to a curve is a line that touches the curve at a single point, sharing the same direction or slope as the curve at that point. It represents the instantaneous rate of change of the function and provides a linear approximation of the function near the contact point.

Derivative

The derivative of a function measures the rate at which the function's value changes with respect to changes in the input. It is the primary tool for determining the slope of the tangent line at any given point on the curve.

Condition of Tangency

The condition of tangency requires that the line has a slope equal to the derivative of the curve at the point of contact, and that it intersects the curve at that point. This condition results in an equation that relates the coordinates of the point of tangency with the given slope and the function.

Equation of a Line

The equation of a line can be expressed in forms such as point-slope or slope-intercept form. These forms allow for utilizing known information about the line's slope and a point on the line to construct the full equation describing the line.

Transcript

00:01 So they want us to find all the tangent lines to this equation here with slope negative 1.

00:09 So remember, if this is supposed to be our tangent slope, what this is really telling us is we want to find all equations or all solutions to where our derivative is going to be equal to negative 1.

00:23 So let's go ahead and find our derivative here.

00:25 And before we do that, i'm going to rewrite this first to be x to the minus or x minus 1 raise to the negative 1.

00:35 And now if we take the derivative of this, we can just use power rule.

00:39 So off on the side, i'll kind of write what the power rule is just to kind of remind us.

00:43 So this is going to be x to the n, and this is equal to n times x.

00:50 Now we subtract one from the power, so it would be n minus.

00:54 Let me write that in a little bit better.

00:59 All right.

00:59 So now, if we were to take the derivative of this, it would be y prime is equal to.

01:06 So we move the negative one out front, subtract one off the power.

01:10 So it would be negative 1, x minus 1.

01:13 Now to the negative 2.

01:14 But also recall that when we have a function inside of the function, we need to use the chain rule.

01:21 So this is going to be d by dx.

01:24 But i'll just write this, f of g of x.

01:27 Here, this is equal to f prime of g of x times g prime of x.

01:36 So then we need to multiply by the derivative on the inside here, so d by d x of x minus 1, and well, that's just going to be 1.

01:46 So we can go ahead and leave that like that.

01:49 And then i'm going to rewrite this to be y prime is equal to negative 1 over x minus 1 squared.

01:57 And so now remember what our goal was was to set this equal to negative 1.

02:04 So now we can go ahead and solve here.

02:08 So we would multiply over by x minus 1 squared, and that would give us negative 1 is equal to negative 1 x minus 1 squared.

02:22 We can multiply each side by negative 1, so those just become positive.

02:28 And then we can go ahead and expand the right -hand side out.

02:31 So that gives us 1 is equal to x squared minus 2x plus 1.

02:38 We can subtract 1 from each side, and those ones cancel out.

02:46 And then we will end up with 0 is equal to x squared minus 2x.

02:50 And now we can go ahead and factor the right -hand side, which is going to give us 0 is equal to x -x -minus 2.

03:01 And then using the zero product property that tells us x is equal to zero or x minus 2 is equal to 0.

03:08 So we can add the 2 over.

03:10 So we have x is equal to zero or x is equal to 2.

03:13 So these are going to be the two solutions that make our derivative negative 1.

03:18 So let me go ahead and write out the equation for the slopes or the equations of the lines here...

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