The number of points, where the curve $$f(x) = e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1$$, $$x \in \mathbb{R}$$ cuts $$x$$-axis, is equal to

To find the points where the curve cuts the x-axis, set

$$e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1=0$$

Let

$$t=e^{2x}$$

Since $$e^{2x}>0$$ for all real $$x$$,

$$t>0$$

The equation becomes

$$t^4-t^3-3t^2-t+1=0$$

Dividing throughout by $$t^2$$,

$$t^2-t-3-\frac{1}{t}+\frac{1}{t^2}=0$$

Rearranging,

$$\left(t^2+\frac{1}{t^2}\right)-\left(t+\frac{1}{t}\right)-3=0$$

Let

$$u=t+\frac{1}{t}$$

Using

$$u^2=t^2+2+\frac{1}{t^2}$$

gives

$$t^2+\frac{1}{t^2}=u^2-2$$

Substituting,

$$(u^2-2)-u-3=0$$

$$u^2-u-5=0$$

Solving,

$$u=\frac{1\pm\sqrt{21}}{2}$$

Since $$t>0$$,

$$u=t+\frac{1}{t}\geq 2$$

Therefore,

$$u=\frac{1+\sqrt{21}}{2}$$

is the only admissible value.

Substituting back,

$$t+\frac{1}{t}=k$$

where

$$k=\frac{1+\sqrt{21}}{2}>2$$

Multiplying by $$t$$,

$$t^2-kt+1=0$$

The discriminant is

$$\Delta=k^2-4$$

Since $$k>2$$,

$$\Delta>0$$

Therefore, the quadratic has two distinct real roots.

Also,

$$t_1t_2=1>0$$

and

$$t_1+t_2=k>0$$

Hence, both roots are positive.

Since

$$t=e^{2x}$$

and the exponential function is one-to-one, each positive value of $$t$$ gives exactly one real value of $$x$$.

Thus, the equation has two distinct real solutions for $$x$$.

Hence, the curve cuts the x-axis at

$$2$$

distinct points.