Find the real values of $P$ for which $f(x)=P$ has exactly one solution.

Question

$$f(x)=(x-1)^2(x-2)+1$$ Find the real values of $P$ for which $f(x)=P$ has exactly one solution.

Hi, I'm a little bit confused with this question.I don't know how I should start. I think I need to use the quadratic formula but I don't know if it gives me the right answer. Can you help me with this question?

Answer 1

We need $(x-1)^2(x-2) = P-1$ to have exactly one real root. The LHS has a double root at $1$ and a single root at $2$, and is increasing otherwise. Hence as long as $P-1$ does not cross the LHS in this interval, it will always have only a real root.

Therefore the only values $P-1$ cannot take is the range of $(x-1)^2(x-2)$ when $x \in [1, 2]$. This is easily found to be $[-\frac4{27}, 0]$, either using calculus or the AM-GM: $$\frac14(x-1)^2(2-x) \le \left( \frac{\frac{x-1}2+\frac{x-1}2+2-x}3\right)^3=\frac1{27}$$

Answer 2

I think it is very useful to think about this exercise graphically. Let us have a look at the following graph.

The graph shows the function $$ f(x)=(x-1)^2(x-2)+1 $$ and a horizontal line $f(x)=1.5$ which depicts the situation when $P=1.5$. We need to find such values $P$ that the horizontal line interesects the graph of the function only once.

Answer 3

Just another way considering that, once expanded, you have to solve the cubic equation $$x^3-4 x^2+5 x-(1+P)=0$$

If you look here, you will see that the key factor for the number of roots of the general cubic equation $ax^3+bx^2+cx+d=0$ is the discriminant $\Delta$ given by $$\Delta=18 a b c d - 4 b^3 d + b^2 c^2 - 4 a c^3 - 27 a^2 d^2$$ So, for the case $a=1$, $b=-4$, $c=5$, $d=-(1+P)$, $$\Delta=-27 P^2+50 P-23=-27(P-1)(P-\frac {23}{27})$$ As given in the Wikipedia page, if $\Delta<0$, the equation has one real root and two nonreal complex conjugate roots.

I am sure that you can take it from here.

Answer 4

1) Find the local minimum and maximum of $f(x)=(x-1)^2(x-2)+1$.

$~~~~~$For this purpose, find $x_1$ and $x_2$ such that $\frac{d}{dx} f(x_1)= \frac{d}{dx} f(x_1) =0$. ($ x_1 = 1$ is a local maximum and $x_2 = \frac{5}{3}$ is a local minimum)

2) For each $P>f(1)=1$ or $P<f(\frac{5}{3})=\frac{23}{27}$, $f(x)=P$ has exactly one solution.

3) For $P=1$ or $P=\frac{23}{27}$, $f(x)=P$ has two solutions.

4) For $P\in(\frac{23}{27}~~,~~1)$ , $f(x)=P$ has three solutions.