Answer
$$-1\lt x \lt \frac{11}{3}$$
Work Step by Step
To have a steeper tangent line, we must have $$\frac{dy}{dx}=8x+11> \frac{dy}{dx}=3x^2.$$
Now, we have the inequality
$$8x+11>3x^2\Longrightarrow 3x^2-8x-11<0\Longrightarrow 3x^2+3x-11x-11<0$$
which leads to
$$3x(x+1)-11(x+1)<0\Longrightarrow (x+1)(3x-11)<0.$$
Then, we have the solution
$$-1\lt x \lt \frac{11}{3}$$
