using Rolles theorem to see if it can be applied to f. If so, find all numbers c such that f`(c)=0 f(x)=sin(x)+cos(x) interval [0,2pi]
since the period of sin and cos is 2pi, the function clearly satisfies Rolle's Theorem.
So, we want to find c (possibly more than one value) where
f'(c) = 0
f'(x) = cosx - sinx
so, f'(c) = 0 if cosx = sinx
That is when x = pi/4 or 5pi/4
See the graph at
http://www.wolframalpha.com/input/?i=sinx%2Bcosx
To determine if Rolle's theorem can be applied to the function f(x) = sin(x) + cos(x) on the interval [0, 2π], we need to check the following conditions:
1. Continuity: The function f(x) should be continuous on the closed interval [0, 2π].
2. Differentiability: The function f(x) should be differentiable on the open interval (0, 2π).
3. Equal Function Values: The function f(x) should have the same function values at the endpoints of the interval.
Let's check each condition:
1. Continuity: The sine and cosine functions are continuous everywhere, so the sum sin(x) + cos(x) is also continuous everywhere, including the interval [0, 2π]. Condition satisfied.
2. Differentiability: Both sin(x) and cos(x) are differentiable everywhere, so their sum sin(x) + cos(x) is also differentiable everywhere, including the open interval (0, 2π). Condition satisfied.
3. Equal Function Values: We need to evaluate f(0) and f(2π) to check if they are equal.
- f(0) = sin(0) + cos(0) = 0 + 1 = 1
- f(2π) = sin(2π) + cos(2π) = 0 + 1 = 1
Since f(0) = f(2π) = 1, the third condition is satisfied.
Thus, all the conditions for Rolle's theorem are satisfied for the function f(x) = sin(x) + cos(x) on the interval [0, 2π].
To find all numbers c such that f'(c) = 0, we need to find the derivative of f(x) and solve for f'(x) = 0.
First, let's find the derivative of f(x):
f'(x) = d/dx (sin(x) + cos(x))
= cos(x) - sin(x)
Next, we solve for f'(x) = 0:
cos(x) - sin(x) = 0
To find the values of x where cos(x) - sin(x) = 0, we can rearrange the equation:
cos(x) = sin(x)
From the unit circle, we know that sin(x) = cos(π/2 - x). Substituting this in, we get:
cos(x) = cos(π/2 - x)
To find the values of x that satisfy this equation, we equate the arguments of the cosine functions:
x = π/2 - x
Simplifying,
2x = π/2
Therefore, x = π/4.
So, the number c such that f'(c) = 0 is c = π/4.