Analyze the following functions. Which function has relative extrema?
f(x)=4x-6.7
f(x)=cos(x-3)+5
f(x)= √ (50x)
f(x)=1.4tan(x+2)
To find the relative extrema of each function, we need to determine the critical points by taking the derivative of the function and setting it equal to zero.
1. f(x)=4x-6.7
f'(x) = 4
Setting f'(x) = 0 gives us no critical points, so there are no relative extrema in this function.
2. f(x)=cos(x-3)+5
f'(x) = -sin(x-3)
Setting f'(x) = 0 gives us sin(x-3) = 0. This happens when x-3 = n*π, where n is an integer. So potential critical points are x = 3 + n*π. Plugging these values into f(x) to determine if they are relative extrema.
3. f(x)= √ (50x)
f'(x) = 25/(√50x)
Setting f'(x) = 0 gives us no critical points, so there are no relative extrema in this function.
4. f(x)=1.4tan(x+2)
f'(x) = 1.4sec^2(x+2)
Setting f'(x) = 0 gives us sec^2(x+2) = 0. This does not happen as secant squared function cannot be zero, so there are no critical points and no relative extrema in this function.
In conclusion, the only function with relative extrema is f(x)=cos(x-3)+5.