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Given the function $f(x) = \dfrac{(x-1)}x$, find the equation of the tangent line to the graph of $f$ that pass though the point $(4,1)$. NOTE: The point (4,1) is NOT on the graph of f.

Okay so first I found the slope of the tangent through the following:

$m=f'(x)=(f'(x)g(x)-f(x)g'(x))/(g(x))^2$

$= [(1)(x)-(x-1)(1)]/ x^2$

$= (x-(x-1))/x^2$

$= 1/x^2$ So now what do I do?

Hamou
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stacks-geek
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2 Answers2

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So we have the slope of our tangent line to be the slope of our function.

$$f'(x) = \frac{1}{x^2}$$

At some point on $f(x)$ and our tangent, which we can denote as $(c, \frac{c-1}{c})$

Our slope now becomes:

$$f'(c) = \frac{1}{c^2}$$

Now we have a point for our tangent line, $(4, 1)$ and another point $(c, \frac{c-1}{c})$

So let us use our point-slope form:

$$y - 1 = m(x-1)$$ $$y - \frac{c-1}{c} = \frac{1}{c^2}(x-c)$$ Now we can use our point $(4, 1)$

$$1 - \frac{c-1}{c} = \frac{1}{c^2}(4-c)$$

Solving we find that $c = 2$

So, our tangent line is

$$y - \frac{2-1}{2} = \frac{1}{4}(x-2)$$

$$y =\frac{x}{4}$$

Varun Iyer
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Hint

The point-slope equation of a tangent line at a point $(x_0,f(x_0))$ is $$y-f(x_0)=f'(x_0)(x-x_0).$$ If the line goes trhough the point $(4,1)$ then you have to solve the equation $$4-f(x_0)=f'(x_0)(1-x_0)$$

mfl
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