Question

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 local maximum, local minimum, or point of inflection.

Answer 1

\[\int\limits x(x+2)(x-5)dx=\frac{x^4}{4}-x^3-5 x^2 \]A plot of f(x) is attached.

Answer 2

since f' is given, the critical numbers are x=0, -2, 5. Do the first derivative test to see where the function is increasing/decreasing on (-inf, -2), (-2, 0), (0, 5), and (5, inf) to find local max/min.
To find inflection points, find f'' and set to 0.

Answer 3

Set the second derivative to zero\[3 x^2-6 x-10=0 \]and solve for x, the two inflection points.\[\left\{x\to \frac{1}{3} \left(3-\sqrt{39}\right),x\to \frac{1}{3} \left(3+\sqrt{39}\right)\right\} \]or\[\{x\to -1.08167,x\to 3.08167\} \]Confirm these points with the plot.