determine all inflection points of f(x)=3x^5 + 5x^4 - 50x^3 + 90x^2 + x - 4. as your answer, please enter…

Explanation:

Step1: Find the first - derivative

Using the power rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=15x^{4}+20x^{3}-150x^{2}+180x + 1$.

Step2: Find the second - derivative

Differentiate $f'(x)$ again. $f''(x)=60x^{3}+60x^{2}-300x + 180$.

Step3: Set the second - derivative equal to zero

Factor out 60 from $f''(x)$: $60(x^{3}+x^{2}-5x + 3)=0$, so $x^{3}+x^{2}-5x + 3 = 0$. By trial and error, we find that $x = 1$ is a root. Then we perform polynomial long - division: $(x^{3}+x^{2}-5x + 3)\div(x - 1)=x^{2}+2x - 3$. Factor $x^{2}+2x - 3=(x + 3)(x - 1)$. So, $x^{3}+x^{2}-5x + 3=(x - 1)^{2}(x+3)=0$. The solutions are $x=-3$ and $x = 1$.

Step4: Check for concavity change

We take test points in the intervals $(-\infty,-3)$, $(-3,1)$ and $(1,\infty)$. For $x=-4$: $f''(-4)=60\times(-4)^{3}+60\times(-4)^{2}-300\times(-4)+180=60\times(-64)+60\times16 + 1200+180=-3840+960+1200 + 180=-1500<0$. For $x = 0$: $f''(0)=180>0$. For $x = 2$: $f''(2)=60\times2^{3}+60\times2^{2}-300\times2+180=480+240-600 + 180=300>0$. The concavity changes at $x=-3$. The point $x = 1$ is not an inflection point since the concavity does not change at $x = 1$.

Answer:

-3