Q. 2 - Calculus

Question

Global extrema and derivatives: The first-derivative test can be used to show that the function $f\left(x\right) = x^3-x^2+x$ has a local maximum at $x=\frac{1}{3}$ and a local minimum at x = 1. Are either of these local extrema also global extrema? Can the first or second derivative tell you whether or not a local extremum is a global extremum?

Step-by-Step Solution

Answer

Given function doesn't have any local extrema. 

Step 1. Given Information.

Given a function: $f\left(x\right) = x^3-x^2+x.$

Step 2. Firstly find the critical points of the function.

To find the critical points of the function we need to solve $f\text{'}\left(x\right) = 0 for x.$

So,

$$\begin{gathered} f\left(x\right) = x^3-x^2+x \\ f\text{'}\left(x\right) = 3x^2-2x+1 = 0, \\ Solving the quadratic by finding the discrimant first, \\ we get, D = b^2-4ac = 4-4.3.1=4-12=-8. \\ Since the discriminant is negative so the derivative has no real roots. \\ So, the given function has not any critical points. \\ So, the function does not have any local extremum. \end{gathered}$$