The function $f$ is such that $f(x) = 2x^3 + 5x^2 - 4x - 3$, where $x \in \mathbb{R}.$ (a) Show that $x = -3$ is a root of $f(x)$ and find the other two roots - Leaving Cert Mathematics - Question 5 - 2017

To find the local extrema of the function, we first calculate its derivative:

$$y = 2x^3 + 5x^2 - 4x - 3 \ \frac{dy}{dx} = 6x^2 + 10x - 4.$$

Setting the derivative to zero to find critical points:

$$6x^2 + 10x - 4 = 0$$

Using the quadratic formula:

$$x = \frac{-10 \pm \sqrt{10^2 - 4(6)(-4)}}{2(6)} = \frac{-10 \pm \sqrt{100 + 96}}{12} = \frac{-10 \pm \sqrt{196}}{12} = \frac{-10 \pm 14}{12}.$$

This gives:

1. $x = \frac{4}{12} = \frac{1}{3}$
2. $x = \frac{-24}{12} = -2$.

Next, we calculate the function values at these critical points:

For $x = \frac{1}{3}$:

$$f\left(\frac{1}{3}\right) = 2\left(\frac{1}{3}\right)^3 + 5\left(\frac{1}{3}\right)^2 - 4\left(\frac{1}{3}\right) - 3 = 2\left(\frac{1}{27}\right) + 5\left(\frac{1}{9}\right) - \frac{4}{3} - 3.$$

Simplifying the above gives:

$$= \frac{2}{27} + \frac{15}{27} - \frac{36}{27} - \frac{81}{27} = \frac{-100}{27},$$

Thus, the local maximum is at $\left( -2, \frac{-100}{27} \right)$.

For $x = -2$:

$$f(-2) = 2(-2)^3 + 5(-2)^2 - 4(-2) - 3 = -16 + 20 + 8 - 3 = 9.$$

Thus, the local minimum is at $(-2, 9)$.

For the equation $f(x) + a = 0$ to have only one real root, the function $f(x)$ must touch the x-axis at one point. This occurs when the discriminant of the quadratic is zero:

$b^2 - 4ac = 0.$

Given $f(x) = 2x^3 + 5x^2 - 4x - 3$, we observe that in order to determine the range of $a$, we must consider the maximum and minimum points previously determined:

• Maximum point value is $\frac{-100}{27}$.
• Minimum point value is $9$.

Therefore, for $f(x) + a$ to only touch the x-axis:

$a > \frac{100}{27} \text{ or } a < -9.$

Hence, the range of possible values for $a$ is:

$a > \frac{100}{27} \text{ or } a < -9.$