Question

If $f\left(x\right)=\frac{80}{3x^4+8x^3-18x^2+60}$, then the points of local maxima for the function $f\left(x\right)$ are

• A. $1,3$
• B. $-3,1$
• C. $-1,3$
• D. $-1,-3$

Answer 1

The correct option is B $-3,1$

We have,
$f\left(x\right)=\frac{80}{3x^4+8x^3-18x^2+60}$
$\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{-80(12x^3+24x^2-36x)}{{(3x^4+8x^3-18x^2+60)}^2}$
$=\frac{(-80)\left(12\right)\left(x\right)(x^2+2x-3)}{{(3x^4+8x^3-18x^2+60)}^2}$
$=\frac{-\left(80\right)\left(12\right)\left(x\right)(x-1)(x+3)}{{(3x^4+8x^3-18x^2+60)}^2}$
Put $f^{{}^{\prime }}\left(x\right)=0$
$\Rightarrow x=0,x=1$ and $x=-3$
Clearly, the sign scheme of $f^{{}^{\prime }}\left(x\right)$ is (fig)
Hence, $x=-3$ and $x=1$ are the points of maxima.