Show that the equation
\[
(x-1)^{4}+(x+1)^{4}=c
\]
has exactly two real roots if \(c>2,\) one root if \(c=2\) and no roots
if \(c<2\).
How many real roots does the equation \(\left(x-3\right)^{4}+\left(x-1\right)^{4}=c\)
have?
How many real roots does the equation \(\left|x-3\right|+\left|x-1\right|=c\)
have?
How many real roots does the equation \(\left(x-3\right)^{3}+\left(x-1\right)^{3}=c\)
have?
{[}The answers to parts (ii), (iii) and (iv)
may depend on the value of \(c\). You should give reasons for your
answers.{]}
Solution:
\(\,\)
\begin{align*}
&& c &= (x-1)^4+(x+1)^4 \\
&&&= 2x^4+12x^2+2 \\
\Rightarrow && 0 &= (x^2+6)^2-\frac{c}{2} - 35 \\
\Rightarrow && \underbrace{x^2+6}_{\geq 6} &= \pm \sqrt{35 + \frac{c}{2}}\\
\end{align*}
Therefore there are two solutions if \(c > 2\), one solution if \(c = 2\) and no solutions otherwise.
\(\,\)
This equation is the same equation if \(y = x-2\), ie there are two solutions if \(c > 2\), one solution if \(c = 2\) and no solutions otherwise.
Rewriting as \(|x-1|+|x+1| = c\) we have
For \(x < -1\): \(1-x-1-x = -2x\)
For \(-1 \leq x \leq 1\): \(1-x+x+1 = 2\)
For \(x > 1\): \(x-1+x+1 = 2x\)
Therefore there are infinitely many solutions if \(c = 2\) (the interval \([-3,-1]\)), two solutions if \(c > 2\) and none otherwise.
Rewriting as \((x-1)^3+(x+1)^3\) we have \(x^3+6x = c\). Notice that \(3x^2+6 > 0\) so the function is increasing, ie there is one solution for all \(c\)