Respuesta :
[tex]\begin{gathered} -\infty\:
1) We need to proceed with some tests to find the inflection points of f(x). So, let's take the second derivative of this:
[tex]\begin{gathered} f(x)=x^3-3x^2-5x+7 \\ \\ \end{gathered}[/tex]Following this procedure:
[tex]\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}[/tex]2) Let's take the derivatives using the power rule:
[tex]\begin{gathered} \frac{d^2}{dx^2}\left(x^3-3x^2-5x+7\right) \\ \\ \frac{d}{dx}\left(x^3-3x^2-5x+7\right)=3x^2-6x-5 \\ \\ \frac{d}{dx}\left(3x^2-6x-5\right)=6x-6 \end{gathered}[/tex]3) Now, let's work for the inflection point f''(x)=0
[tex]\begin{gathered} f\:^{\doubleprime}\left(x\right)=0 \\ \\ 6x-6=0 \\ \\ 6x=6 \\ \\ \frac{6x}{6}=\frac{6}{6} \\ \\ x=1\Rightarrow(1,0) \end{gathered}[/tex]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.
[tex]\begin{gathered} f^{\prime}^{\prime}(x)>0 \\ \\ 6x-6>0 \\ \\ x>1\:\:\:\:Concavity\:upwards \\ \\ --- \\ 6x-6<0 \\ \\ 6x<6 \\ \\ \frac{6x}{6}<\frac{6}{6} \\ \\ x<1\:\:\:\:Concavity\:downwards \end{gathered}[/tex]