Generate a 2D graphical representation of the mathematical function f(x) = (e^x)*cos(x), with the domain extending from 0 to two pi. The graph should be neat and clear, emphasizing the peaks and troughs of the function, hinting to the viewer about the potential points of maximum, minimum and inflection.

2. Consider the function f defined by f(x)=(e^X)cosx with domain[0,2pie] .

a. Find the absolute maximum and minimum values of f(x)
b. Find the intervals on which f is increasing.
c. Find the x-coordinate of each point of inflection of the graph of f.

f(x) = e^x (cosx)

f'(x) = e^x(-sinx) + e^x(cosx)
= e^x(cosx - sinx) = 0 for max/min values of f(x)
so e^x = 0 , no solution
or
cosx - sinx = 0
sinx= cosx
sinx/cosx = 1
tanx = 1
x = π/4 or 5π/4

f(0) = e^0(cos0) = 1
f(2π) = e^(2π)(1) = e^(2π) = appr. 535.5
f(π/4) = e^(π/4) (√2/2) = appr. 1.55
f(5π/4) = e^(5π/4) cos 5π/4 = appr. -35.9

take it from there

b) the function is increasing when f'(x) > 0
e^x(cosx - sinx) > 0
since e^x > 0 for all x
this results in cosx - sinx >0
-sinx > -cosx
sinx/cosx < 1
tanx < 1
So for the domain from 0 to 2π
tanx < 1 for
0 < x < π/4 OR π/2 < x < 5π/4 OR 3π/2 < x < 2π
( I looked at the tangent curve for these)

c) take the derivative of f'(x)
f''(x) = e^x(-cosx) + e^x(-sinx) + e^x(-sinx) + e^x(cosx)
= -2e^x sinx
= 0 for pts of inflection

then sinx = 0
x = 0 , π, 2π

a. To find the absolute maximum and minimum values of the function f(x) = (e^x)cos(x) in the given domain [0, 2π], we first need to find the critical points and the endpoints of the interval.

1. Critical points:
To find the critical points, we need to find the values of x where the derivative of f(x) equals zero or is undefined.

f'(x) = d/dx[(e^x)cos(x)]
= (e^x)(-sin(x)) + (e^x)(cos(x))
= (e^x)(cos(x) - sin(x))

Setting f'(x) = 0:
(e^x)(cos(x) - sin(x)) = 0

For (e^x)(cos(x) - sin(x)) = 0, either (e^x) = 0 or (cos(x) - sin(x)) = 0.

(e^x) = 0 has no solution in the given domain [0, 2π].

(cos(x) - sin(x)) = 0:
cos(x) = sin(x)
Dividing both sides by cos(x):
1 = tan(x)

Solving for x, we get x = π/4.

Therefore, the critical point is x = π/4.

2. Endpoints:
The endpoints of the given domain [0, 2π] are x = 0 and x = 2π.

Now, we need to evaluate the function f(x) at the critical point and the endpoints.

f(0) = (e^0)cos(0) = 1 * 1 = 1
f(2π) = (e^(2π))cos(2π) = 1 * 1 = 1
f(π/4) = (e^(π/4))cos(π/4)

Using a calculator to approximate (e^(π/4)) ≈ 1.284 and cos(π/4) = √2/2:

f(π/4) ≈ 1.284 * (√2/2) ≈ 0.910

Therefore, the absolute maximum value is 1 (at x = 0 and x = 2π) and the absolute minimum value is approximately 0.910 (at x = π/4).

b. To find the intervals on which f(x) is increasing, we need to analyze the derivative of f(x).

As we found earlier, f'(x) = (e^x)(cos(x) - sin(x)).

Since f'(x) = 0 at x = π/4, and f'(x) does not change sign at x = π/4 (from negative to positive), f(x) is not increasing around x = π/4.

Therefore, f(x) is increasing on the interval [0, π/4) and (π/4, 2π].

c. To find the x-coordinate of each point of inflection of the graph of f, we need to find the values of x where the second derivative of f(x) changes sign.

f''(x) = d^2/dx^2[(e^x)cos(x)]
= [(e^x)(-sin(x)) + (e^x)(cos(x))(-sin(x))] + [(e^x)(-sin(x)) - (e^x)(cos(x))(cos(x))]
= (e^x)(-2sin(x)cos(x))

Setting f''(x) = 0:
(e^x)(-2sin(x)cos(x)) = 0

For (e^x)(-2sin(x)cos(x)) = 0, either (e^x) = 0 or (-2sin(x)cos(x)) = 0.

(e^x) = 0 has no solution in the given domain [0, 2π].

(-2sin(x)cos(x)) = 0:
sin(x) = 0 or cos(x) = 0

For sin(x) = 0, x can be 0, π, or 2π.

For cos(x) = 0, x can be π/2 or 3π/2.

Therefore, the x-coordinates of the points of inflection are x = 0, x = π/2, x = π, x = 3π/2, and x = 2π.

To find the absolute maximum and minimum values of the function f(x) = e^x * cos(x) on the interval [0, 2π], we can follow these steps:

a. To find the critical points, we need to find where the derivative of f(x) is equal to 0 or does not exist.

First, let's find the derivative of f(x) using the product rule. The derivative of e^x is e^x, and the derivative of cos(x) is -sin(x).

f'(x) = e^x * cos(x) - e^x * sin(x)

Next, we set f'(x) equal to 0 and solve for x:

e^x * cos(x) - e^x * sin(x) = 0

Factoring out e^x:

e^x * (cos(x) - sin(x)) = 0

This equation is satisfied when either e^x = 0 or cos(x) - sin(x) = 0.

However, e^x is never equal to 0, so we focus on the second term:

cos(x) - sin(x) = 0

Rearranging the equation:

cos(x) = sin(x)

This equation is satisfied when x = π/4 or x = 5π/4.

b. To determine the intervals on which f(x) is increasing, we need to examine the sign of the derivative f'(x) in different regions.

We can use a sign chart or check the sign of f'(x) at some test points.

For x < π/4:
Choose a test point, like x = 0. Plug it into f'(x) = e^x * cos(x) - e^x * sin(x).
f'(0) = 1 * cos(0) - 1 * sin(0) = 1 > 0

For π/4 < x < 5π/4:
Choose a test point, like x = π.
f'(π) = e^π * cos(π) - e^π * sin(π) = -1 - 0 = -1 < 0

For x > 5π/4:
Choose a test point, like x = 2π.
f'(2π) = e^(2π) * cos(2π) - e^(2π) * sin(2π) = 1 - 0 = 1 > 0

From this analysis, we can conclude that f(x) is increasing on the interval [0, π/4) and (5π/4, 2π].

c. To find the x-coordinate of each point of inflection of the graph of f, we need to find where the second derivative of f(x) changes sign or is equal to 0.

First, let's find the second derivative of f(x):

f''(x) = (e^x * cos(x) - e^x * sin(x))' = (e^x * cos(x) - e^x * sin(x))' = e^x * cos(x) - e^x * sin(x) - e^x * sin(x) - e^x * cos(x) = -2e^x * sin(x)

The second derivative is -2e^x * sin(x).

To find the points of inflection, we need to find where the second derivative is equal to 0 or changes sign.

Setting -2e^x * sin(x) = 0, we get sin(x) = 0. This is satisfied at points where x is an integer multiple of π (e.g., x = 0, x = π, x = 2π, etc.).

Therefore, the x-coordinates of the points of inflection of the graph of f are x = nπ, where n is an integer.

sorry for the double post