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Question: How do you find the slope and equation of tangent line to \[f(x) = - \dfrac{1}{x}\]at\[\left( {3, - ...

How do you find the slope and equation of tangent line to f(x)=1xf(x) = - \dfrac{1}{x}at(3,13)\left( {3, - \dfrac{1}{3}} \right)?

Explanation

Solution

In-order to find slope, you need to find f(x)f'(x) and then the value of f(x)f'(x) at the given point will give you slope of the tangent line at given point. Once the slope is known you can use the point slope formula to find the equation of the tangent line.

Formula Used: Derivative of xn{x^n} which will be n×xn1n \times {x^{n - 1}} and also we need to use slope-point line equation formula which is yy1=m(xx1)y - {y_1} = m(x - {x_1}) in-order to find the tangent equation at the point.

Complete step by step solution:
Given the function,
f(x)=1xf(x) = - \dfrac{1}{x}
We need to find the slope of tangent at the given point(3,13)\left( {3, - \dfrac{1}{3}} \right). So we will find the derivative of the function first.
dydxf(x)=dydx(1x)\dfrac{{dy}}{{dx}}f(x) = \dfrac{{dy}}{{dx}}\left( { - \dfrac{1}{x}} \right)
We can write 1x - \dfrac{1}{x} as (x1) - \left( {{x^{ - 1}}} \right)
As we know that derivative of xn{x^n} will be n×xn1n \times {x^{n - 1}} using that, we will get
f1(x)=1x2\Rightarrow {f^1}(x) = \dfrac{1}{{{x^2}}}
Now substituting the value of x at the point (3,13)\left( {3, - \dfrac{1}{3}} \right) which is 3.
We will get
f1(3)=132{f^1}(3) = \dfrac{1}{{{3^2}}}
f1(3)=19\Rightarrow {f^1}(3) = \dfrac{1}{9}
So the slope of the tangent line at the given point will be m=19m = \dfrac{1}{9}
Using the point-slope formula (equation) to find the tangent equation at the given point (3,13)\left( {3, - \dfrac{1}{3}} \right)
yy1=m(xx1)\Rightarrow y - {y_1} = m(x - {x_1})
On Substituting the point and slope we will get
y(13)=19(x3)\Rightarrow y - \left( { - \dfrac{1}{3}} \right) = \dfrac{1}{9}\left( {x - 3} \right)
On simplifying the equation we will get
y+13=19(x3)\Rightarrow y + \dfrac{1}{3} = \dfrac{1}{9}(x - 3)
Multiplying with 9 on both sides we will get

9y+9×13=x3 9y+3=x3 9yx+6=0  \Rightarrow 9y + 9 \times \dfrac{1}{3} = x - 3 \\\ \Rightarrow 9y + 3 = x - 3 \\\ \Rightarrow 9y - x + 6 = 0 \\\

Therefore the tangent equation for the function at the given point is 9yx+6=09y - x + 6 = 0 and slope ism=19m = \dfrac{1}{9}.

Note: When graph functions are given, the derivative will be the slope of the tangent line at the given point. Slope of a line is the change in vertical axis with the change of horizontal axis whereas derivation of a function is also the change of one variable with respect to the other variable so we can say the derivation of a function also gives the slope of the line.