condition for a cubic equation to have no roots in an interval

Question

I want to find the possible values of a parameter $a$ such that the equation $2x^3-9x^2+30-a=0$ does not have a root in $(-2,2)$. I am stuck can anyone help

Answer 1

Plot $f(x)=2x^3-9x^2+30$. The critical points are the roots of $f'(x)=0$ and are easy to find.

The critical points are $x=0$ and $x=3$. We have $f(0)=30$ and $f(3)=3$. The graph shows that the equation $f(x)=a$ has a single solution for $a<3$ or $a>30$. For $a=3$ or $a=30$, the equation has two solutions (a double one). For $3 < a < 30$, the equation has three solutions. Now argue about $x \in (-2,2)$.

Answer 2

Hint: Write your equation in the form $$2x^3-9x^2+20=a$$ and consider the function $$f(x)=2x^3-9x^2+30$$ $$f'(x)=6x(x-3)$$ $$f''(x)=6(2x-3)$$

Answer 3

Differentiate the function $$f(x)=2x^3-9x^2+30-a$$ to get $$f'(x) = 6x(x-3)$$

Thus your critical points are $x=0$ and $x=3$

On the interval $-(-2, 2)$ the critical point $ (0,30-a)$ is a maximum point.

Therefore the function f(x) does not have any zeros in the interval $(-2,2)$ if and only if $$30-a <0$$ that is if and only if $$a>30$$