Question

find the inflection points of (f(x)=3x^{4}+41x^{3}-21x^{2}+13). (give your answers as a comma separated list, e.g., 3,-2). inflection points =

Explanation:

Step1: Find the first - derivative

Using the power rule $(x^n)'=nx^{n - 1}$, if $f(x)=3x^{4}+41x^{3}-21x^{2}+13$, then $f'(x)=12x^{3}+123x^{2}-42x$.

Step2: Find the second - derivative

Differentiate $f'(x)$ again. $f''(x)=36x^{2}+246x - 42$.

Step3: Set the second - derivative equal to zero

$36x^{2}+246x - 42 = 0$. Divide through by 6 to simplify: $6x^{2}+41x - 7=0$.

Step4: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 6$, $b = 41$, $c=-7$), use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(41)^{2}-4\times6\times(-7)=1681 + 168=1849$. Then $x=\frac{-41\pm\sqrt{1849}}{12}=\frac{-41\pm43}{12}$.

Step5: Find the x - values of the inflection points

$x_1=\frac{-41 + 43}{12}=\frac{2}{12}=\frac{1}{6}$ and $x_2=\frac{-41-43}{12}=\frac{-84}{12}=-7$.

Answer:

$\frac{1}{6},-7$