Analyze the following functions. Which function has relative extrema? f(x) = 1.4tan(x + 2); f(x) = 4x - 6.7; f(x) = sqrt(50x); f(x) = sin(x - 3) + 5
To find the relative extrema of a function, we need to find the critical points by setting the derivative equal to zero, then determine if the function has a maximum or minimum at those points.
1. f(x) = 1.4tan(x + 2)
Taking the derivative of f(x):
f'(x) = 1.4sec^2(x + 2)
Setting f'(x) = 0:
1.4sec^2(x + 2) = 0
There are no solutions to this equation, so there are no critical points and thus no relative extrema for this function.
2. f(x) = 4x - 6.7
Taking the derivative of f(x):
f'(x) = 4
Setting f'(x) = 0:
4 = 0
This equation has no solutions since 4 is never equal to 0. Therefore, there are no critical points and no relative extrema for this function.
3. f(x) = sqrt(50x)
Taking the derivative of f(x):
f'(x) = 25 / sqrt(50x)
Setting f'(x) = 0:
25 / sqrt(50x) = 0
This equation also has no solutions since the square root term cannot be zero. Therefore, there are no critical points and no relative extrema for this function.
4. f(x) = sin(x - 3) + 5
Taking the derivative of f(x):
f'(x) = cos(x - 3)
Setting f'(x) = 0:
cos(x - 3) = 0
x - 3 = pi/2 or x - 3 = 3pi/2
Solving for x, we get x = pi/2 + 3 or x = 3pi/2 + 3
These are the critical points of the function. To determine if they are relative extrema, we look at the sign of the second derivative:
f''(x) = -sin(x - 3)
At x = pi/2 + 3, f''(pi/2 + 3) = -sin(pi/2) = -1 < 0
Therefore, x = pi/2 + 3 is a local maximum.
At x = 3pi/2 + 3, f''(3pi/2 + 3) = -sin(3pi/2) = 1 > 0
Therefore, x = 3pi/2 + 3 is a local minimum.
In conclusion, the function f(x) = sin(x - 3) + 5 has relative extrema at x = pi/2 + 3 (local maximum) and x = 3pi/2 + 3 (local minimum).