 Coby Allison

2022-02-15

What is the equation of the line tangent to $f\left(x\right)=4\mathrm{sec}x-8\mathrm{cos}x$ at x=3? Arif Coates

Equation of line tangent can be formed by ${y}_{1}-y=m\left({x}_{1}-x\right)$ where m is the slope and ${x}_{1}$ and ${y}_{1}$ represent the x-coordinate and y-coordinate of the point of intersection of f(x) and tangent.
We know that slope of line tangent to y=f(x) is ${f}^{\prime }\left(x\right)=\frac{dy}{dx}$.
So, to do this question, we have to find f'(x) first.
$f\left(x\right)=4\mathrm{sec}x-8\mathrm{cos}x$
${f}^{\prime }\left(x\right)=4\mathrm{sec}x\mathrm{tan}x-8\left(-\mathrm{sin}x\right)$
$=4\mathrm{sec}x\mathrm{tan}x+8\mathrm{sin}x$
When x=3, the slope of line tangent (m)=f'(3)
$=4\mathrm{sec}\left(3\right)\mathrm{tan}\left(3\right)+8\mathrm{sin}\left(3\right)$
and the y-coordinate of this point is f(3)
$=4\mathrm{sec}\left(3\right)-8\mathrm{cos}\left(3\right)$
We get all the information we need and we can plug them into the equation.
Equation of line tangent at x=3:
$4\mathrm{sec}\left(3\right)-8\mathrm{cos}\left(3\right)-y=\left[4\mathrm{sec}\left(3\right)\mathrm{tan}\left(3\right)+8\mathrm{sin}\left(3\right)\right]\left(3-x\right)$

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