Application of second derivative
In calculus, we know that if we want to find the slope of a tangent at any given point on the graph of a function, we find the derivative of that function. This derivative that we find is called the first derivative of that function. The first derivative of the function besides giving us the slope of the tangent at a particular point, also gives us the critical point. When we equate the first derivative f’(x) of a function y = f(x), to zero and then solve for x, we get some values of x. These values of x are called critical points. They are denoted by c. These critical points divide the x axis into various intervals. Next we use the first derivative to find the intervals in which the function is increasing and the intervals in which the function is decreasing. At the critical points, f’(x) = 0, therefore at these points the function would have either a local maximum or a local minimum. The point at which the slope of a tangent to the curve is 0 is called the critical point. Thus at the critical point, the tangent would be horizontal. Therefore summarizing we can say that,
Critical point occurs when f’(x) = 0 or f’(x) does not exist.
Function f(x) is increasing in the interval where f’(x) > 0 and it is decreasing in the interval where f’(x) < 0.
That sums up how we use the first derivative for understanding the graph of a function. Now let us take a look at how the second derivative also helps us to understand the graph of a function.
Just like how the first derivative helps us to know where a function is increasing or decreasing, the second derivative helps us to know where a function is concave up and where is it concave down. When f’’(x) > 0, the function in that interval is concave up and if f”(x) < 0, then in that interval the function is concave down. The point at which the concavity of the function changes from concave up to concave down or from concave down to concave up is called the inflection point. As in the case of first derivative, when we equated f’(x) to zero and solved for x we got the critical point. In the same way, in case of the second derivative, when we equate f”(x) to zero and solve for x we get the x co ordinate of the point of inflection. To find the y co ordinate of the point we substitute the x co ordinate thus found into the original function f(x).
Consider the following example.
Find the inflection point for the function f(x) = x^4 + 2x^3 – 12x^2
Solution: First we find f’(x) for the given f(x). Therefore,
f(x) = x^4 + 2x^3 – 12x^2
f’(x) = 4x^3 + 6x^2 – 24x
Next we find the second derivative. Thus we get,
f”(x) = 12x^2 + 12x – 24
Now we set the f”(x) = 0. So,
12x^2 + 12x – 24 = 0
X^2 + x – 2 = 0
(x+2)(x-1) = 0
X = -2, x = 1
(-2, -48), (1, -9)