The number of points where the curve f(x)=e^8x-e^6x-3e^4x-e^2x+1, x R cuts the x-axis, is …

Question

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

Answer: 2

Accepted Answer

Let $t=e^{2x}>0$. The equation becomes $t^4-t^3-3t^2-t+1=0$. Dividing by $t^2$:
$$\left(t^2+\frac1{t^2}\right)-\left(t+\frac1t\right)-3=0.$$

Let $p=t+\dfrac1t$ (so $t^2+\dfrac1{t^2}=p^2-2$):
$$p^2-2-p-3=0 \implies p^2-p-5=0 \implies p=\frac{1\pm\sqrt{21}}2.$$

Since $t>0$, $p\ge2$; only $p=\dfrac{1+\sqrt{21}}2\approx2.79$ qualifies.

Then $t^2-pt+1=0$ has discriminant $p^2-4\approx3.78>0$, and since both the sum ($p>0$) and product ($1$) of roots are positive, both roots of $t$ are positive real numbers. Each gives one value $x=\dfrac{\ln t}2$.

So there are 2 real solutions, i.e. the curve cuts the $x$-axis at 2 points.

Solutions · 1