The derivative of a function is f'(x)=x(x+2)(x-5). Find the value of x at each point where f has a
(a) local maximum,
(b) local minimum, or
(c) point of inflection
Set f'=0
x=0,-2,5 for f' to be zero.
maximum is when f" is negative
a) f"=(x+2)(x-5)+x(x-5)+(x(x+2)
at x=0, f"=-10 so x=0 is max
b. f" at x=-2
f"=(0+ -2(-7) = positive, so at x=-2, it is minimum
c. f" at x=5
f"=0+0+35 positive, local maximum
check my calcs
oops, when f" is 35, local min
To find the values of x at each point where f has a local maximum, local minimum, or point of inflection, we need to analyze the behavior of the derivative.
(a) Local Maximum:
A local maximum occurs when the derivative changes from positive to negative. In other words, it is where the slope of the function changes from increasing to decreasing. To find the x-values at local maxima, we need to find the critical points by setting the derivative equal to zero and solving for x.
f'(x) = x(x+2)(x-5) = 0
To find the critical points, we must solve this equation. We can use the zero product property, which states that if a product of factors is equal to zero, then at least one of the factors must be zero. Therefore, we set each factor equal to zero and solve for x individually.
x = 0 (factor 1)
x + 2 = 0 (factor 2)
x - 5 = 0 (factor 3)
Solving these equations individually, we find:
x = 0, -2, and 5
Therefore, f has local maxima at x = 0, x = -2, and x = 5.
(b) Local Minimum:
A local minimum occurs when the derivative changes from negative to positive. In other words, it is where the slope of the function changes from decreasing to increasing. To find the x-values at local minima, we again need to find the critical points by setting the derivative equal to zero and solving for x.
f'(x) = x(x+2)(x-5) = 0
Setting each factor equal to zero and solving for x, we already found x = 0, -2, and 5 as the critical points. These critical points can also be local minima if the function changes from decreasing to increasing at those points.
Therefore, f has local minima at x = 0, x = -2, and x = 5.
(c) Point of Inflection:
A point of inflection occurs where the concavity of the function changes. To find the x-values at points of inflection, we need to check the second derivative. The second derivative is the derivative of the derivative, or the derivative of f'(x).
f''(x) = (x(x+2)(x-5))' = (3x^2 - 11x - 10)'
Differentiating the expression, we find:
f''(x) = 6x - 11
To find the points of inflection, we set this second derivative equal to zero and solve for x.
6x - 11 = 0
Solving the equation, we find:
x = 11/6
Therefore, f has a point of inflection at x = 11/6.