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

Beginner2022-02-16Added 6 answers

Equation of line tangent can be formed by ${y}_{1}-y=m({x}_{1}-x)$ 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.

$f\left(x\right)=4\mathrm{sec}x-8\mathrm{cos}x$

${f}^{\prime}\left(x\right)=4\mathrm{sec}x\mathrm{tan}x-8(-\mathrm{sin}x)$

$=4\mathrm{sec}x\mathrm{tan}x+8\mathrm{sin}x$

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:

$4\mathrm{sec}\left(3\right)-8\mathrm{cos}\left(3\right)-y=[4\mathrm{sec}\left(3\right)\mathrm{tan}\left(3\right)+8\mathrm{sin}\left(3\right)](3-x)$

We know that slope of line tangent to y=f(x) is

So, to do this question, we have to find f'(x) first.

When x=3, the slope of line tangent (m)=f'(3)

and the y-coordinate of this point is f(3)

We get all the information we need and we can plug them into the equation.

Equation of line tangent at x=3: