The Second derivative Test for concavity can be proved using Mean Value Theorem.

**Given:** f(x) is a continuous function in an interval [a, b] and f"(x) > 0 in (a, b)

**Prove:** f(x) is concave up in the interval (a, b).

Let c be any number in (a,b). To prove f(x) is concave up it has to be shown that the curve y = f(x) lies

above the tangent at the point (c, f(c)).

Since the equation to the tangent line is y = f(c) + f'(c) (x-c) it has to be shown f(x) > f(c) + f'(c) (x-c) for all x ∈ (a,b) and x ≠ c.

Let us first prove this for the case x > c.

Hence by mean value theorem we can find a number 'k' in (c, x) i.e c < k < x such that

f(x) - f(c) = f'(k) (x -c)

⇒ f(x) = f(c) + f'(k) (x -c) ------------------------------(1)

It is given f"(x) > 0 in (a, b) which implies f'(x) is increasing in (a, b)

We have both c and k ∈ (a,b) and c < k

Hence f'(c) < f'(k)

⇒ f'(c) (x -c) < f'(k) (x -c) x-c is positive as x > c

Adding f(c) to both the sides of the inequality,f(c) + f'(c)( x-c) < f(c) + f'(k) (x-c)

f(c) + f'(c) (x-c) < f(x) Substitution using (1)

or

f(x) > f(c) + f'(c) (x -c)

which proves that the curve y = f(x) lies above the tangent line at any point in the given interval.

Case (2) x < c

By mean value theorem, we can find a number 'k' in (x, c) such that

f(c) - f(x) = f'(k) (c-x)

⇒ f(x) = f(c) + f'(k) (x-c) ---------------------------------------------(2)

As f'(x) is increasing in (a, b)

f'(c) > f'(k)

f'(c) (x-c) < f'(k) (x -c) x-c is negative. The inequality is reversed.

Adding f(c) to both the sides of the inequality,f(c) + f'(c) (x-c) < f(c) + f'(k) (x-c)

f(c) + f'(c) (x-c) < f(x) Substitution using (2)

or

f(x) > f(c) + f'(c) (x-c)

which is the statement required.

The second derivative test for the case f"(x) < 0 can also be proved, using similar steps.