Question

PLEASE HELP WITH THIS CALCULUS PROBLEM!



a) Find an equation of a line that is tangent to the curve f(x) = 2cos3x and whose slope is a maximum.

b) Is this the only possible solution? Explain. If there are other possible solutions, how many solutions are there?

PLEASE SHOW ALL WORK!!

Answer 1

take the second derivative of f. find x such that f " = 0... these will be the max/min of your first derivative (slope of your tangent line). choose the x that gives you a max.

since f is periodic, there are other solutions.

Answer 2

Maximizing the slope means maximizing the derivative.
\[f'(x)=-6\sin{3x}\]
The maximum value of f'(x) is 6 which means that sin(3x)=-1 which means x=(4npi-pi)/6 for integer n. For our convenience, let's pick n=0, so that x=-pi/6.
The tangent line at x=-pi/6 is:
\[y=6(x+\frac{\pi}{6})+f(\frac{-\pi}{6})\]
And as noted, there are infinitely other solutions.

Answer 3

so the tangent of the line is what u wrote in that last line?

Answer 4

im confused...

Answer 5

That's the equation we're looking for.

Answer 6

how do i show my work to be able to derive that?

Answer 7

Just do what I did and before the last equation, use the fact that the equation of the tangent line at a point (a, f(a)) is \(\large y=f'(a)(x-a)+f(a)\) where a=-pi/6 in this case.

Answer 8

ok thanks :)

Answer 9

No prob. :)