To determine which function has relative extrema, we need to find the critical points by setting the derivative of each function equal to zero and solve for x.
1. For f(x) = 4x - 6.7, the derivative is f'(x) = 4. Setting this equal to zero gives us 4 = 0, which has no solution.
2. For f(x) = sin(x - 3) + 5, the derivative is f'(x) = cos(x - 3). Setting this equal to zero gives us cos(x - 3) = 0. This occurs when x - 3 = π/2, 3π/2, etc. Therefore, there are relative extrema at x = π/2 + 3, 3π/2 + 3, etc.
3. For f(x) = √50x, the derivative is f'(x) = √50. Setting this equal to zero gives us √50 = 0, which has no solution.
4. For f(x) = 1.4tan(x + 2), the derivative is f'(x) = 1.4sec^2(x + 2). Setting this equal to zero gives us sec^2(x + 2) = 0, which has no real solution.
Therefore, the function f(x) = sin(x - 3) + 5 has relative extrema at x = π/2 + 3, 3π/2 + 3, etc.
Analyze the following functions. Which function has relative extrema?(1 point) Responses f(x)=4x−6.7 f left parenthesis x right parenthesis equals 4 x minus 6.7 f(x)=sin(x−3)+5 f left parenthesis x right parenthesis equals sine left parenthesis x minus 3 right parenthesis plus 5 f(x)=50x−−−√ f left parenthesis x right parenthesis equals Start Root 50 x End Root f(x)=1.4tan(x+2)
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