The function y=2x^3-3ax^2 + 6 is decreasing only on the interval (0,3). Find a?

Sep 10, 2015

The answer is $a = 3$.

Explanation:

If the function is decreasing on the interval (0, 3), that means its derivative must be negative on this interval. So first let's calculate the derivative, using the power rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} 2 {x}^{3} - 3 a {x}^{2} + 6 = 6 {x}^{2} - 6 a x$

So we know that the derivative $6 {x}^{2} - 6 a x$ is negative on the interval $\left(0 , 3\right)$ and positive everywhere else. That means this function must cross the $x$ axis at 0 and at 3. Or in other words, when $x = 0$ or when $x = 3$, the function must be equal to 0. We can use this information to solve backwards for $a$.

No matter what $a$ is, when $x = 0$, the function will be 0, because $6 \cdot {0}^{2} - 6 \cdot a \cdot 0 = 0$. So that doesn't help us.

But we also know that when $x = 3$, the function also has to be 0, so we can write:
$6 \cdot 9 - 6 \cdot 3 a = 0$
$54 = 18 a$
$a = 3$

To double check this, we can plug in $a = 3$ and graph the function $y = 2 {x}^{3} - 9 {x}^{2} + 6$:

graph{2x^3 - 9x^2 + 6 [-3, 5, -26.6, 13.4]}

And we see that it is decreasing on the interval (0, 3).