I would please like your guidance to find, depending on $a$, how many roots the following equation has: $x^4+4x+a=0$, where $a$ is a real parameter.
I tried to use Bolzano's Theorem for a specific interval but I am afraid that my approach would not be accurate. Thank you very much in advance.
Hint. Sketch the graph of $x^4 + 4x$. Adding the constant $a$ just shifts that graph vertically.
Hint:
Compute the derivative of $p(x)=x^4+4x+a$ and prove this polynomial first decreases from $+\infty$ to a single minimum, then increases to $+\infty$. Therefore $p(x)$ has $2,1$ or no root according to whether $\min p(x)<, =\:$ or $>0$.
Let $f(x) = x^4 + 4x + a$ then
$f'(x) = 4x^3 + 4$
$f''(x)= 12x^2$.
$f'(x) =0$ if $x^3 +1=0$ or if $x=-1$ so and $f''(1)=12 > 0$. so $f(x)$ has a minimum at $x =-1$ and $f(-1) = 1-4+a = a-3$.
If $a > 3$ then $f(-1) > 0$ and the minimum of $f$ is where $f(x) = a-3> 0$ and $f(x) =0$ never occurs. So $f(x) > 0$ and has no real roots if $x > 3$.
If $a = 3$ then $f(-1)=a-3 = 0$ and the minimum of $f$ is where $f(x) = a-3=0$ and that is the only time $f(x) = 0$ and there is only one real root.
If $a < 3$ then $f(-1)=a-3 < 0$ and the minimum of $f(x) < 0$. But as $x\to \pm \infty$, $f(x)\to \infty$ there must be points where $x < -1$ and $x > -1$ where $f(x)$ is positive and by intermediate value theorem,, and thus points where $f(x) = 0$. And $x=-1$ is the only extrema, there are only two such real roots.
$x^4=-4x -a$;
Note: A secant line intersects the convex curve $y=x^4$ in at most $2$ points. Hence at most $2$ real solutions to the problem.
Consider the intersection of
$y=x^4$ with the line $y=-4x-a$.
1) $a \le 0$.
The secant line $y=-4x -a$ intersects the convex curve $y=x^4$ in two points.
2) $a>0$.
Tangent line:
Find $a$ s.t. $y=-4x-a$ is a tangent to $y=x^4$.
$4x^3=-4$; $x=-1$;
Tangent passes through $x=-1$, $y=1$:
$1=4-a$; $a=3$;
Hence
$1$ solution for $a=3$;
$ 2$ solutions for $0 \le a <3$.
4) No solutions for $a >3$.
