Review question

# When does $3x^4-16x^3+18x^2=k$ have exactly two real roots? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6537

## Solution

Find the coordinates of each turning point on the graph of $y=3x^4-16x^3+18x^2$ and determine in each case whether it is a maximum point or a minimum point.

We find that $\dfrac{dy}{dx}=12x^3-48x^2+36x=12x(x^2-4x+3)=12x(x-1)(x-3)$.

Since turning points occur where $\dfrac{dy}{dx}=0$, the turning points are at $x=0,1$ or $3$.

When $x=0$ we have that $y=0$, when $x=1$ we find that $y=5$, and when $x=3$, $y=-27$.

Therefore the turning points are $(0,0),$ $(1,5)$ and $(3,-27)$.

We can consider $y'=\dfrac{dy}{dx}$ at different values of $x$ to identify if a point is a minimum or a maximum.

At $x=-\dfrac{1}{2}$ we observe that $\dfrac{dy}{dx}<0$ and, at $x=\dfrac{1}{2}, \dfrac{dy}{dx}>0$. This shows us that $(0,0)$ is a minimum point.

At $x=\dfrac{3}{2}$ we can see that $\dfrac{dy}{dx}>0$, so considering the result above at $x=\dfrac{1}{2}$ this shows that $(1,5)$ is a maximum.

It is evident that for all $x>3$, $\dfrac{dy}{dx}>0$, implying that $(3,-27)$ is a minimum point.

Or else we could say that $y'' = 36x^2-96x+36 = 12(3x^2-8x+3)$.

So $y''(0)$ is positive, $y''(1)$ is negative and $y''(3)$ is positive.

This means $(0,0),$ is a minimum, $(1,5)$ is a maximum, and $(3,-27)$ is a minimum.

Sketch the graph of $y=3x^4-16x^3+18x^2$, and state the set of values of $k$ for which the equation $3x^4-16x^3+18x^2=k$ has precisely two real roots for $x$.

From the information that we have, we can sketch the graph of $y=3x^4-16x^3+18x^2$ as follows.

By considering the values of $k$ for which there are exactly two intersections between the graph we have just sketched and $y=k$, we find that the values required are $-27<k<0, \quad \text{and} \quad 5<k.$