Step 1:
Concept
An inflection point is a point on the graph of a function at which the concavity changes. Points of inflection can occur where the second derivative is zero. In other words, solve f '' = 0 to find the potential inflection points. Even if f ''(c) = 0, you can't conclude that there is an inflection at x = c.
Step 2:
Find the first derivative of the function.
[tex]\begin{gathered} f(x)=5x^3-30x^2\text{ - 7x - 2} \\ f^{\prime}(x)=15x^2\text{ - 60x - 7} \end{gathered}[/tex]Step 3:
Find the second derivative
[tex]\begin{gathered} f^{\prime}(x)=15x^2\text{ - 60x - 7} \\ f^{\doubleprime}(x)\text{ = 30x - 60} \end{gathered}[/tex]Step 4:
Next, set the second derivative equal to zero and solve. Because f '' is a polynomial, we have to factor it to find its roots.
[tex]\begin{gathered} 30x\text{ - 60 = 0} \\ 30x\text{ = 60} \\ \text{x = }\frac{60}{30} \\ \text{x = 2} \end{gathered}[/tex]The solution is x = 2
Final answer
[tex]\begin{gathered} f(x)=5x^3-30x^2-7x\text{ - 2} \\ =5(2)^3-30(2)^2-7(2)\text{ - 2} \\ \text{= 40 - 120 - 14 - 2} \\ =\text{ }-96 \end{gathered}[/tex]Inflection point = (2 , -96)