Find the x value for all maxima and minima of f(x) if:
f'(x)=(x-7)(x+3)(x-8)^2(x+27)^3
remember that max/mins occur when f ' (x) equal zero.
Since your f ' (x) is already nicely factored, all we have to do is set each of the factors equal to zero and solve, so
the x's that yield a max/min are
x = 7, -3, 8, and -27
thank you
To find the x values for all maxima and minima of f(x), we need to find the critical points of the function. Critical points occur where the derivative of the function equals zero or is undefined.
In this case, the derivative of f(x) is given by f'(x) = (x-7)(x+3)(x-8)^2(x+27)^3. To find the critical points, we set f'(x) equal to zero and solve for x.
Setting f'(x) = 0, we have:
(x-7)(x+3)(x-8)^2(x+27)^3 = 0
Now, we need to solve this equation to find the values of x that make this expression equal to zero. To do this, we can use the zero product property, which states that if a product of factors equals zero, then at least one of the factors must be zero.
Therefore, we set each factor equal to zero:
x - 7 = 0 --> x = 7
x + 3 = 0 --> x = -3
(x - 8)^2 = 0 --> x - 8 = 0 --> x = 8
(x + 27)^3 = 0 --> x + 27 = 0 --> x = -27
So, the critical points of f(x) are x = 7, x = -3, x = 8, and x = -27.
These values of x correspond to the locations of the maxima and minima of the function f(x). To determine whether they are maxima or minima, we can examine the behavior of the function around these points or use the second derivative test.