Final answer:
To find the x-values of relative extrema of the function f(x)=x³+3x²-9x-1, calculate the derivative, set it to zero, solve the resulting quadratic equation, and then substitute these x-values into f(x) to find their corresponding y-values. Use the Second Derivative Test or the First Derivative Test to confirm their nature as maxima or minima.
Step-by-step explanation:
To find the x-values where the function f(x)=x³+3x²-9x-1 has any relative extrema, we first need to find its derivative to identify potential critical points. The derivative f'(x) represents the slope of f(x) at any given point:
f'(x) = 3x² + 6x - 9
We then set the derivative equal to zero to find critical points where the slope of the graph is zero (where potential maxima or minima occur):
3x² + 6x - 9 = 0
This is a quadratic equation which can be factored or solved using the quadratic formula. The quadratic formula for an equation of the form ax²+bx+c=0 is:
x = ∛(-b ± √(b²-4ac))/(2a)
Substituting in our coefficients a=3, b=6, and c=-9 we find two solutions for x, which are the x-values of the relative extrema. These values of x can then be substituted back into the original function f(x) to find the y-values of the relative extrema.
Once the potential critical points are found, the Second Derivative Test or the First Derivative Test can be used to determine if these points are indeed maxima or minima.