The maximum interval in which the slopes of t | vedclass

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(D) Let $f(x) = y = x^4+5x^3+9x^2+6x+2$. The slope of the tangent to the curve is given by $m(x) = \frac{dy}{dx} = 4x^3+15x^2+18x+6$. For the slope $m

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$R-\left(\frac{-3}{2}, -1\right)$

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(D) Let $f(x) = y = x^4+5x^3+9x^2+6x+2$. The slope of the tangent to the curve is given by $m(x) = \frac{dy}{dx} = 4x^3+15x^2+18x+6$. For the slope $m(x)$ to be increasing, its derivative must be greater than or equal to zero, i.e.,$\frac{dm}{dx} = \frac{d^2y}{dx^2} \ge 0$. Calculating the second derivative: $\frac{d^2y}{dx^2} = 12x^2+30x+18$. Factoring the quadratic expression: $12x^2+30x+18 = 6(2x^2+5x+3) = 6(2x+3)(x+1)$. We need $6(2x+3)(x+1) \ge 0$. The critical points are $x = -\frac{3}{2}$ and $x = -1$. Testing the intervals: $1$) For $x \in (-\infty, -\frac{3}{2}]$, $\frac{d^2y}{dx^2} \ge 0$. $2$) For $x \in [-\frac{3}{2}, -1]$, $\frac{d^2y}{dx^2} \le 0$. $3$) For $x \in [-1, \infty)$, $\frac{d^2y}{dx^2} \ge 0$. Thus, the slope increases in the interval $(-\infty, -\frac{3}{2}] \cup [-1, \infty)$, which can be written as $R - (-\frac{3}{2}, -1)$.

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

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