Question

how do i solve f(x)=sec x, (pi/3,2)

Answer 1

What do you mean solve f(x)=sec(x), (pi/3,2)?

Answer 2

its calculus.. product and quotient rules and higher order derivatives

Answer 3

I still don't have any instructions what to with it f(x)=sec(x), (pi/3,2)
Does the instructions say to differentiate f(x)=sec(x) at (pi/3,2)
Does the instructions say to find the tangent line to f(x)=sec(x) at (pi/3,2)

Answer 4

how to solve f(x)=sec(x), (pi/3,2) gives no instructions to anyone

Answer 5

in fact you cannot solve f(x)=sec(x) , (pi/3,2)

Answer 6

you can solve 2=sec(x) for x
and one value of x satisfying that is x=pi/3

Answer 7

Assuming you want to differentiate f(x)=sec(x) at (pi/3,2)
You would first to need to know (sec(x))'=sec(x)tan(x)

Answer 8

i apologize, the directions are to find the tangent line to that problem

Answer 9

Ok lines have the form
y-y_1=m(x-x_1)
where (x_1,y_1) is a point you know on the line
And since we are talking about tangent lines at (pi/3,2)
we know (x_1,y_1) is (pi/3,2) since it is on the line
m is the slope
in order to find the slope you must first find f'(x)
then evaluate f'(pi/3)

Answer 10

so your equation will look like this:
\[y-2=f'(\frac{\pi}{3})(x-\frac{\pi}{3})\]

Answer 11

you just have to find the f'(pi/3) part

Answer 12

so i just have to find the slope of the for the equation ? im sorry im just really lost in calc.

Answer 13

yes and slopes can be found from the derivative

Answer 14

all right

Answer 15

so you need to know the derivative of sec(x)

Answer 16

which i will give you since it should be something you commit to memory anyways
(sec(x))'=sec(x)tan(x)

Answer 17

now you just replace the x's in sec(x)tan(x) with pi/3
and evaluate

Answer 18

use the unit circle

Answer 19

thank you !