Using the Product Rule differentiate the function?

Question

f(x)= (cos x) e^-x sqrt 3

Then find stationary points of the function f(x), that lie within the interval [-pi, pi] using the first derivative test to classify (and specify to 4 signif figs) each stationary point as a local max &/or min of f(x).

Any ideas?

Answer 1

f(x) = √3 cos x e^(-x)

Taking the derivative:

f'(x) = √3 d(cos x)/dx e^(-x) + √3 cos x d(e^(-x))/dx
f'(x) = - √3 sin x e^(-x) - √3 cos x e^(-x)

The stationary points are precisely where f'(x)=0, so setting it equal to zero and solving:

0 = - √3 sin x e^(-x) - √3 cos x e^(-x)

-√3 e^(-x) is never equal to zero, so we may divide by it:

0 = sin x + cos x

Now, subtract cos x:

sin x = -cos x

Divide by cos x:

tan x = -1

The points where tan x = -1 in [-π, π] are -π/4 and 3π/4. To determine whether these are local maxima or minima we must determine whether the first derivative is increasing or decreasing at each critical point. Since the first derivative is only zero at two points in the interval [-π, π], it suffices to find the sign of the derivative in the intervals [-π, -π/4), (-π/4, 3π/4), and (3π/4, π]. Recall that the first derivative is:

- √3 sin x e^(-x) - √3 cos x e^(-x)

Since dividing it by any positive number will not change the sign, we can actually factor out √3 e^(-x), as it is always positive. So sgn(f'(x)) = sgn(-sin x - cos x) = -sgn (sin x + cos x). Now we find points to test in each of the intervals. To make the math easier, we test it at -π/2, 0, and π. This yields:

On [-π, -π/4), sgn (f'(x)) = -sgn (sin (-π/2) + cos (-π/2)) = -sgn (-1) = 1, so f is increasing on this interval.

On (-π/4, 3π/4), sgn(f'(x)) = -sgn (sin 0 + cos 0) = -sgn (1) = -1, so f is decreasing on this interval.

On (3π/4, π], sgn (f'(x)) = -sgn (sin π + cos π) = -sgn (-1) = 1, so f is increasing on this interval.

This means that -π/4 is a local maximum, and 3π/4 is a local minimum.

Note: I really don't know why your teacher wants you to compute the first derivative to four significant digits. As I explained earlier, only the sign of the first derivative between those points matters for classifying local maxima and minima, the actual value does not. Still, you may want to actually plug the values into f'(x) on your calculator and write down the numerical answer -- some teachers are really picky about following directions to the letter.

Answer 2

start up with the 4. 4 equals 2 squared besides as 2+2. then you definately take the three/2. 3 halves is what ex-better halves will take in case you get married and seperated three times. So 3/2 & the 4 provides sixteen. stable success.

Answer 3

f(x)' = - sin(x)*e^(-x*sqrt(3)) - sqrt(3)*cox(x)*e^(-x*sqrt(3)).
At stationary points f(x)' = 0 => tg(x) = - sqrt(3) => x = - pi/3 or 2*pi/3.
To classify you need the value of the second derivative at stationary points.
f''(x) = - cos(x)*e^(-x*sqrt(3)) + sqrt(3)*sin(x)*e^(-x*sqrt(3))
+ sqrt(3)*sin(x)*e^(-x*sqrt(3)) + 3*cos(x)*e^(-x*sqrt(3))
= 2*e^(-x*sqrt(3))*(sin(x) + cos(x)).
f''(- pi/3) = -4.4901 < 0 therefore at x = -pi/3 f(-pi/3) = 3.0668 (maximum).
f''(2*pi/3) = 0.0194 > 0 therefore at x = 2*pi/3 f(2*pi/3) = -0.0132 (minimum).