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I am stuck at this. Any help will be much appreciated.

Given $f(x)=\frac{2}{x}$,

  1. Draw the graph of $f$.
  2. Find the equation of the tangent to the graph of $f$ at the given point $M(2,f(2))$ and draw its graph.
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    Where are you stuck? – 2017-01-02
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    Start by calculating $f'(2)$, which gives the slope of the line. Then you will have the slope and a point on the line, which is enough to find the equation. – 2017-01-02
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    https://www.youtube.com/watch?v=TJoy8nHKwRc this link can help you. – 2017-01-02

4 Answers 4

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Well, I'll leave it to you to draw the graph. Notice the derivative:

$$f'(2)=\lim_{h\to0}\frac{f(2+h)-f(2)}h=\lim_{h\to0}\frac{\frac2{2+h}-1}h=\lim_{h\to0}\frac{-h}{h(2+h)}=\lim_{h\to0}\frac{-1}{2+h}=-\frac12$$

And equation of a line tangent to a point $(2,f(2))$ is given by

$$y-f(2)=f'(2)(x-2)$$

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    Thanks so much I could not think of the right equation! Thank you for your help! – 2017-01-02
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    @EleonoraDiamantiJaniSali No problem! Glad to help :D – 2017-01-02
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differentiate to get $$f'(x)=2lnx$$ then use the general equation of a line: $$\frac{y-y(1)}{x-x(1)}=(slope.of .tangent)$$

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    You integrated. – 2017-03-22
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First, we evaluate the derivative $f'(2)$. $$f(x)=\frac{2}{x}$$ Hence, $$f'(x)=-\frac{2}{x^2}$$ Therefore: $$f'(2)=-\frac{2}{4}=-\frac{1}{2}$$

Hence, the gradient of the tangent at $x=2$ is $m=-\frac{1}{2}$.

The equation of a straight line (the tangent in this case) is given by $y=mx+c$.

Substitute the values of $m$, then substitute $y$ and $x$ of your point $(2, f(2))$ to evaluate the value of $c$ to align your tangent to the curve.

Hence,

$$f(2)=-\frac{1}{2} \times 2+c \tag{1}$$

Evaluating $f(2)$:

$$f(2)=\frac{2}{2}=1$$

Hence, we evaluate $c$ by rearranging equation $(1)$.

$$1=-1+c$$ $$c=2$$

Hence, the equation of your tangent is:

$$y=-\frac{1}{2} x+2$$

In an alternative form:

$$x+2y=4$$

I assume you know how to plot graphs.

If you have any doubts or questions, please do not hesitate to ask.

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    yes i can draw the graph its just that for the past few days I 've studied so much algebra that my brain is fried! thanks so much! – 2017-01-02
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tangent

Goal

Given the function $$ f(x) = \frac{2}{x} $$ and the value $x_{*} = 2$, find $$ y = mx + b, $$ the equation of the line tangent to $f(x)$ at $x_{*} = 2$.

Strategy

  1. The slope of the tangent line is given by $m = f'(x_{*})$.
  2. Use the slope-intercept equation $f(x_{*}) = m x_{*} + b$ to find $b$.

Slope $m$

Compute the derivative function $$ f'(x) = -\frac{2}{x^2}. $$ Compute the slope at this location $$ f'(x_{*}) = -\frac{2}{x_{*}^2} = -\frac{1}{2} $$

Intercept $b$

Use the slope intercept form $y=mx+b$ to find $b$. We can use any point $(x,f(x))$, but we already have the point $(x,f(x))$. $$ f(x_{*}) = m x_{*} + b \qquad \Rightarrow \qquad b = 2 $$

Solution $$ y = mx + b = -\frac{1}{2} x + 3 $$