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

Question

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

Answer: 2

Accepted Answer

Firstly, we know that the given function

$f(x)={e}^{8x}-{e}^{6x}-3{e}^{4x}-{e}^{2x}+1$ intersects the x-axis

where $f(x)=0$. Setting $f(x)$ equal to zero gives us :

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

Let $t={e}^{2x}$. The equation now becomes :

${t}^{4}-{t}^{3}-3{t}^{2}-t+1=0.$

Dividing by ${t}^{2}$ and rearranging the equation gives :

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

If we let $y=t+\frac{1}{t}$, we get :

${y}^{2}-y-5=0.$

This quadratic equation in y can be solved using the quadratic formula to give two roots :

$y=\frac{1\pm \sqrt{21}}{2}.$

Since $y=t+\frac{1}{t}$, and $t>0$ (as $t={e}^{2x}$), we must choose the root where $y>2$. Thus, we take $y=\frac{1+\sqrt{21}}{2}$.

So, we have :

$t+\frac{1}{t}=\frac{1+\sqrt{21}}{2}.$

Solving for $t$ gives us :

${t}^{2}-yt+1=0,$

or

$t=\frac{y\pm \sqrt{{y}^{2}-4}}{2}.$

Substituting $y=\frac{1+\sqrt{21}}{2}$ into the formula gives :

$t=\frac{\frac{1+\sqrt{21}}{2}\pm \sqrt{{(\frac{1+\sqrt{21}}{2})}^{2}-4}}{2}.$

This quadratic formula for $t$ yields two possible values (t = 2.37 and t = 0.42). Both of them are positive, thus both are acceptable values for $t={e}^{2x}$.

Finally, to get $x$, take the natural log of both sides and divide by 2 :

$x=\frac{1}{2}\ln (t).$

This gives us two solutions for $x$, corresponding to the two positive solutions for $t$. So, the number of points where the curve $f(x)={e}^{8x}-{e}^{6x}-3{e}^{4x}-{e}^{2x}+1$ intersects the x-axis is 2.

Solutions · 1