Answer is: option4

Solution:

Concavity is determined by the second derivative \( y'' \)

First, we compute the first derivative \( y' \):

\[ y' = \frac{d}{dx} (3x^5 - 40x^3 - 21x) \] \[ y' = 15x^4 - 120x^2 - 21 \]

Now, differentiate again to get the second derivative:

\[ y'' = \frac{d}{dx} (15x^4 - 120x^2 - 21) \] \[ y'' = 60x^3 - 240x \] \[ y'' = 60x(x^2 - 4) \]

Setting \( y'' = 0 \):

\[ 60x(x^2 - 4) = 0 \] \[ 60x(x - 2)(x + 2) = 0 \]

This gives critical points:

\[ x = 0, \quad x = 2, \quad x = -2 \]

Sign Analysis of \( y'' \) in different intervals:

1. For \( x < -2 \) (e.g., \( x = -3 \)): \[ y''(-3) = 60(-3)((-3)^2 - 4) = 60(-3)(9 - 4) = 60(-3)(5) = -900 \] Negative, so concave down.
2. For \( -2 < x < 0 \) (e.g., \( x = -1 \)): \[ y''(-1) = 60(-1)((-1)^2 - 4) = 60(-1)(1 - 4) = 60(-1)(-3) = 180 \] Positive, so concave up.
3. For \( 0 < x < 2 \) (e.g., \( x = 1 \)): \[ y''(1) = 60(1)((1)^2 - 4) = 60(1)(1 - 4) = 60(1)(-3) = -180 \] Negative, so concave down.
4. For \( x > 2 \) (e.g., \( x = 3 \)): \[ y''(3) = 60(3)((3)^2 - 4) = 60(3)(9 - 4) = 60(3)(5) = 900 \] Positive, so concave up.

The function is concave up where \( y'' > 0 \), which occurs in the intervals:

\[ (-2, 0) \quad \text{and} \quad (2, \infty) \]

Final Answer: \( -2 < x < 0 \) or \( x > 2 \).