How do you find all critical point and determine the min, max and inflection given f(x)=3x^2-4x+1?

Dec 18, 2016

Critical point x=$\frac{2}{3}$
Minima $- \frac{1}{3}$
No maxima, no inflection point.

Explanation:

Find f'(x) = 6x-4. Equate it to 0 to get $x = \frac{2}{3}$. This is the only critical point.

Now find f"(x) =6, which is a positive number (irrespective of any x) . Hence, there is a minima at x=2/3. The minima would be $3 {\left(\frac{2}{3}\right)}^{2} - 4 \left(\frac{2}{3}\right) + 1$

= $- \frac{1}{3}$

Maxima does not exist.

Since f"(x) $\ne$ 0, there is no inflection point.