Question

(a) find an equation of the tangent line to
the graph of at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of the graphing utility to confirm your
results.

Answer 1

function: y= cos 3x
point: (pi/4, -sqrt2/2)

Answer 2

where wer u stuck ?

Answer 3

i saw u doing a similar problem the other day ? :)

Answer 4

take the derivative

Answer 5

I just get confused when trig is added in

Answer 6

do you know what the derivative is

Answer 7

dy/dx

Answer 8

-3sin3x

Answer 9

right

Answer 10

yes

Answer 11

so, slope = m = dy/dx = -3sin3x

Answer 12

since u want the slope at
point: (pi/4, -sqrt2/2)

evaluate slope at x = pi/4

Answer 13

dy/dx at x = pi/4 :-

-3sin(3*pi/4)

= ?

Answer 14

-0.123

Answer 15

@ganeshie8

Answer 16

that doesnt look right

you should get :-
dy/dx at x = pi/4 :-

-3sin(3*pi/4)

= \(\large \frac{-3}{\sqrt{2}}\)

Answer 17

now that u knw slope, m = \(\large \frac{-3}{\sqrt{2}}\)
point: (pi/4, -sqrt2/2)

wats the equation of tangent in point slope form ?

Answer 18

y-(-sqrt2/3)=(-3/sqrt2)(x-(pi/4)

Answer 19

?

Answer 20

@hartnn

Answer 21

thats right !

Answer 22

So that's for a? What would the graph look like for b? Do I just put that equation into Geogebra?

Answer 23

@ganeshie8

Answer 24

yup
graph all 3 below :-
1) function: y= cos 3x
2) point: (pi/4, -sqrt2/2)
3) tangent : y-(-sqrt2/\(\color{red}{2}\))=(-3/sqrt2)(x-(pi/4)

Answer 25

u should get exactly like the one attached