Question

The maximum interval in which the slopes of the tangents drawn to the curve \[ y = x^4 + 5x^3 + 9x^2 + 6x + 2 \] increase is:

• A. \( \left[ -\frac{3}{2}, -1 \right] \)
• B. \( \left[ 1, \frac{3}{2} \right] \)
• C. \( R - \left[ 1, \frac{3}{2} \right] \)
• D. \( R - \left[ -\frac{3}{2}, -1 \right] \)
  \

Hint:

To find where the slope is increasing, compute \( y'' \) and determine where it is positive. Solve \( y'' = 0 \) to find critical points and check sign changes.

Answer 1

The Correct Option is D

Step 1: Compute the first derivative \( y' \)
The derivative of the given function: \[ y = x^4 + 5x^3 + 9x^2 + 6x + 2 \] \[ y' = \frac{d}{dx} (x^4 + 5x^3 + 9x^2 + 6x + 2). \] \[ = 4x^3 + 15x^2 + 18x + 6. \] Step 2: Compute the second derivative \( y'' \)
\[ y'' = \frac{d}{dx} (4x^3 + 15x^2 + 18x + 6). \] \[ = 12x^2 + 30x + 18. \] Step 3: Find the critical points for concavity change
To find where the slope of the tangent is increasing, solve: \[ y'' = 0. \] \[ 12x^2 + 30x + 18 = 0. \] Dividing by 6: \[ 2x^2 + 5x + 3 = 0. \] Factorizing: \[ (2x + 3)(x + 1) = 0. \] Solving for \( x \): \[ x = -\frac{3}{2}, \quad x = -1. \] Step 4: Determine the intervals where \( y' \) is increasing
Using a sign test on \( y'' \): - For \( x<-\frac{3}{2} \), \( y''>0 \) (increasing). - For \( -\frac{3}{2}<x<-1 \), \( y''<0 \) (decreasing). - For \( x>-1 \), \( y''>0 \) (increasing). Thus, \( y' \) is increasing outside \( \left[ -\frac{3}{2}, -1 \right] \). Step 5: Conclusion
Thus, the correct answer is: \[ \mathbf{R - \left[ -\frac{3}{2}, -1 \right].} \] \bigskip