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

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Answer 1 · 2016-12-18T16:15:37

Critical point x= $2/3$
Minima $-1/3$
No maxima, no inflection point.

Find f'(x) = 6x-4. Equate it to 0 to get $x=2/3$ . 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 $3(2/3)^2 -4(2/3) +1$

= $-1/3$

Maxima does not exist.

Since f"(x) $!=$ 0, there is no inflection point.

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