Which of the following statements are true?

(A) The function $f(x) = \frac{x^4}{4} - \frac{4}{3}x^3 + \frac{x^2}{2} + 6x$ has 3 critical points.

(B) The function $f(x) = |x| + 3$ has no minimum value.

(C) A local maximum value is always the absolute maximum value.

(D) $f(x) = x^2$ has minima at $x=0$.

Choose the correct answer from the options given below:

Critical points occur where the derivative equals zero or doesn't exist.

For $f(x) = \frac{x^4}{4} - \frac{4}{3}x^3 + \frac{x^2}{2} + 6x$:

$f'(x) = x^3 - 4x^2 + x + 6$

Setting the derivative equal to zero:

$x^3 - 4x^2 + x + 6 = 0$

Testing $x = -1$:

$(-1)^3 - 4(-1)^2 + (-1) + 6 = -1 - 4 - 1 + 6 = 0$ ✓

So $(x + 1)$ is a factor.

$x^3 - 4x^2 + x + 6 = (x + 1)(x^2 - 5x + 6)$

$= (x + 1)(x - 2)(x - 3)$

Critical points: $x = -1, 2, 3$

Statement (A) is TRUE.

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For $f(x) = |x| + 3$:

When $x = 0$: $f(0) = |0| + 3 = 3$

When $x > 0$ or $x < 0$: $|x| > 0$, so $f(x) > 3$

Since $|x| \geq 0$ for all $x$:

$f(x) = |x| + 3 \geq 0 + 3 = 3$

The minimum value is $3$ (occurs at $x = 0$).

Statement (B) is FALSE.

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A function can have multiple local maxima, but only the highest is the absolute maximum.

Counter-example: $f(x) = -x^4 + x^2$ has two local maxima but only one absolute maximum.

Statement (C) is FALSE.

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For $f(x) = x^2$:

$f'(x) = 2x = 0$

$x = 0$

Using the second derivative test:

$f''(x) = 2 > 0$

Since $f''(0) > 0$, there is a minimum at $x = 0$.

Also, $f(0) = 0$ and $f(x) = x^2 \geq 0$ for all $x$.

Statement (D) is TRUE.

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Statements (A) and (D) are TRUE.