Question

Find the interval(s) in which f(x) = sec(x) is convex.

• A. $(-\pi /2,\pi /2)$
• B. $(\pi /2,3\pi /2)$
• C. $(3\pi /2,5\pi /2)$
• D. $(5\pi /2,7\pi /2)$

Answer 1

Answer: (A), (C)

$(3\pi /2,5\pi /2)$

When f(x) is convex, we know that f”(x) will be positive. So, we will find f”(x) to decide the intervals in which the given function is convex.

f(x) = sec(x)

f’(x) = sec(x). tan(x)

In f’(x) we have product of two functions so we’ll apply the product rule for differentiation.

f”(x) = sec(x). tan(x) . tan(x) + sec(x). $\left(\right(sec\left(x\right){)}^2$

f”(x) = sec(x) ( $ta{n}^2$(x) + $se{c}^2$(x) )

We can replace secx with 1/cosx to simplify.

$\to =\frac{1}{\cos x}(\frac{(\sin x{)}^2}{(\cos x{)}^2}+\frac{1}{(\cos x{)}^2})$

After simplifying, this becomes

$f"\left(x\right)=\left(\frac{1+(\sin x{)}^2}{(\cos x{)}^3}\right)-------\left(1\right)$

When f(x) is convex, we know that f”(x) will be positive. In (1), numerator is always positive. Now the sign of the expression depends on the sign of denominator. So, we can say f”(x) will be positive or will be convex whenever $(\cos x{)}^3$is positive. This will be +ve whenever cosx is +ve. So, from the given intervals, we just have to find the intervals in which cosx is positive, which is easy from the graph of cosx

As it is clear from the graph, cosx is positive in the intervals $(-\pi /2,\pi /2)$ and $(3\pi /2,5\pi /2)$ . So the options A and C are correct.