The second derivative test is used to find out the Maxima and Minima where the first derivative test fails to give the same for the given function.

Let us consider a function f defined in the interval I and let \(c\in I\). Let the function be twice differentiable at at c. Then,

**(i) Local Minima: **x= c, is a point of local minima, if \(f'(c) = 0\) and \(f”(c) > 0\). The value of local minima at the given point is f(c).

**(ii) Local Maxima: **x= c, is a point of local maxima, where \(f'(c) = 0\) and \(f”(c) < 0\). The value of local maxima at the given point is f(c).

**(iii) **If in case \(f'(c) = 0\) and \(f”(c) = 0\), the second derivative test fails. Thus we go back to the first derivative test.

**Working rules:**

**(i) **In the given interval in f, find all the critical points.

**(ii) **Calculate the value of the functions at all the points found in step (i) and also at the end points.

**(iii) **From the above step, identify the maximum and minimum value of the function, which are said to be absolute maximum and absolute minimum value of the function.

**Point of Inflection:**

If the value of the function does not change the sign as x increases from c, then c is neither a point of Local Maxima or Minima. This is known as Point of Inflection.

\(f'(x) = 3x^{3}+ 24x^{2} + 45x\) \(= 3x(x^{2}+8x+15) = 3x (x+3)(x+5)\) \(f'(x) = 0\) \(\Rightarrow x = 0\;\; or \;\; x = -3 \;\; or \;\; x = -5\) Critical Points = 0, -3, -5 Now, \(f”(x) = 9x^{2} + 48 x +45\) \(= 3 (3x^{2} + 16 x + 15)\) Now checking the value of functions at all the critical point, we have: \(f”(0) = 45 >0,\) point of local minima. \(f”(-3)= -18\), point of local maxima. \(f”(-5)= 30\),, point of local minima. |