# Second Derivative Test to Find Maxima & Minima

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. |