Question

What are possible inflection points of f(x)=x^3 - 3x^2 -5x + 7? Are thereany actual inflection points of f(x)? What are intervals of concavity?Explain everything in detail.

Answer 1

Following this procedure:

`\begin{gathered} \mathrm{I}f\:f\:^(\doubleprime)\left(x\right)>0\:then\:f\left(x\right)\:concave\mathrm{\:}upwards\mathrm{.} \\ \mathrm{If}\:f\:^(\doubleprime)\left(x\right)<0\:\mathrm{then}\:f\left(x\right)\:concave\mathrm{\:}downwards\mathrm{.} \end{gathered}`

2) Let's take the derivatives using the power rule:

`\begin{gathered} (d^2)/(dx^2)\left(x^3-3x^2-5x+7\right) \\ \\ (d)/(dx)\left(x^3-3x^2-5x+7\right)=3x^2-6x-5 \\ \\ (d)/(dx)\left(3x^2-6x-5\right)=6x-6 \end{gathered}`

3) Now, let's work for the inflection point f''(x)=0

`\begin{gathered} f\:^(\doubleprime)\left(x\right)=0 \\ \\ 6x-6=0 \\ \\ 6x=6 \\ \\ (6x)/(6)=(6)/(6) \\ \\ x=1\Rightarrow(1,0) \end{gathered}`

Thus, the inflection point is at (1,0)

4) For the concavity we need to find the intervals where f''(x)>0 and f''(x)<0.

`\begin{gathered} f^(\prime)^(\prime)(x)>0 \\ \\ 6x-6>0 \\ \\ x>1\:\:\:\:Concavity\:upwards \\ \\ --- \\ 6x-6<0 \\ \\ 6x<6 \\ \\ (6x)/(6)<(6)/(6) \\ \\ x<1\:\:\:\:Concavity\:downwards \end{gathered}`