how many distinct real zeros a function has
$f(x)= x^4+2x^3-2x^2+1$
How many distinct real zeroes does $f$ have?
Is it two because it crosses the $x$-axis twice or am I completely wrong?
$\endgroup$ 12 Answers
$\begingroup$Given $$ f(x) = x^4+2x^3-2x^2+1 $$ you have $$ f'(x) = 4x^3+6x^2-4x = 2x (2x^2+3x-2) = 4x(x+2)(x-1/2). $$ So, by looking at the sign of $f'$, you see that $f$ has an absolute minimum at $x=-2$ a local maximum at $x=0$ and a local minimum at $x=1/2$. At infinity one has $f(x)\to +\infty$ as $x\to \pm \infty$. In $x=-2$ one has $f(-2)=-7 <0$, in $x=0$ one has $f(0)=1>0$ and in $x=1/2$ one has $f(1/2)>0$.
By the intermediate value theorem one has a zero $x_1 \in (-\infty,-2)$ and another one in $(-2,0)$ because the function changes its sign. These zeroes are unique because the function is strictly monotone (hence injective) in these intervals. On $[0,+\infty)$ the function is strictly positive because its minimum value is $f(1/2)$.
$\endgroup$ $\begingroup$You could use its discriminant. See
It is $\Delta=-1456$
$\endgroup$
