The straight lines representing the tangent and normal at the point x = 7,
have slopes that are related to the slope of the function.
Response:
[tex]a) \hspace {0.5 cm}The \ equation \ of \ the \ tangent \ line \ is; \ \displaystyle y = \underline{\frac{2}{7} - \frac{x}{49}}[/tex]
[tex]b) \hspace{0.5 cm} The\ equation \ of \ the \ normal \ line \ is; \ y= \underline{49 \cdot x - 342 \dfrac{6}{7}}[/tex]
[tex]The \ function \ is; \ f(x) = \mathbf{ \dfrac{1}{x}}[/tex]
The slope of the tangent line is given as follows;
[tex]Slope = \dfrac{d}{dx} f(x) = f'(x) = \dfrac{d}{dx} \left(\dfrac{1}{x} \right) = \mathbf{ -\dfrac{1}{x^2}}[/tex]
Which gives;
[tex]The \ slope \ at \ x = 7, \, \mathbf{f'(7) }= -\dfrac{1}{7^2} = -\dfrac{1}{49}[/tex]
[tex]At \ x = 7, \, f(7) = \dfrac{1}{7}[/tex]
The coordinate of the point on the line at x = 7, is therefore;
[tex]\mathbf{\left(7, \, \dfrac{1}{7} \right)}[/tex]
The equation of the tangent line is therefore;
[tex]y - \dfrac{1}{7} = \mathbf{-\dfrac{1}{49} \cdot \left(x - 7\tight)}[/tex]
Which gives;
49·y - 7 = 7 - x
49·y = 14 - x
Which gives;
[tex]\displaystyle y = \frac{14}{49} - \frac{x}{49} = \mathbf{\frac{2}{7} - \frac{x}{49}}[/tex]
b) The slope of the normal line is the negative inverse of the slope of
the tangent line, therefore;
The slope of the normal line = [tex]\mathbf{-\frac{1}{-\frac{1}{49} }}[/tex] = 49
The equation of the normal line is; [tex]y - \dfrac{1}{7} = \mathbf{ 49 \cdot \left(x - 7\tight)}[/tex]
Which gives;
[tex]y= 49 \cdot \left(x - 7\right) + \dfrac{1}{7} = 49 \cdot x - 343 + \dfrac{1}{7} = \mathbf{49 \cdot x - 342 \dfrac{6}{7}}[/tex]
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