If the function y=g(x) representing the slope | vedclass

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(D) The slope of the tangent to the curve $y=3x^4-5x^3-12x^2+18x+3$ is given by the derivative $g(x) = \frac{dy}{dx}$.Calculating the derivative: $g(x

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$R-[-\frac{1}{2}, \frac{4}{3}]$

Vedclass · Answer

(D) The slope of the tangent to the curve $y=3x^4-5x^3-12x^2+18x+3$ is given by the derivative $g(x) = \frac{dy}{dx}$.Calculating the derivative: $g(x) = 12x^3 - 15x^2 - 24x + 18$.For $g(x)$ to be strictly increasing,its derivative $g'(x)$ must be greater than $0$.$g'(x) = 36x^2 - 30x - 24$.Setting $g'(x) > 0$: $36x^2 - 30x - 24 > 0$.Dividing by $6$: $6x^2 - 5x - 4 > 0$.Factoring the quadratic: $(3x-4)(2x+1) > 0$.The roots are $x = \frac{4}{3}$ and $x = -\frac{1}{2}$.The inequality holds for $x \in (-\infty, -\frac{1}{2}) \cup (\frac{4}{3}, \infty)$.This is equivalent to $R - [-\frac{1}{2}, \frac{4}{3}]$.Thus,the correct option is $D$.

If the function $y=g(x)$ representing the slopes of the tangents drawn to the curve $y=3x^4-5x^3-12x^2+18x+3$ is strictly increasing,then the domain of $g(x)$ is:

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