Prove that the logarithmic function is strictly increasing on (0, ∞).

Open in App

Solution

The given function is f( x )=logx.

Differentiate the function with respect to x.

f ′ ( x )= d dx logx = 1 x

It is clear that when x>0,

f ′ ( x )>0 1 x >0

Thus, the function f( x )=logx is strictly increasing in the interval ( 0,∞ ).

Suggest Corrections

Similar questions

Prove that function f given by $f (x) = l o g (c o s x)$ is strictly decreasing on $(0, \frac{π}{2})$ and strictly increasing on $(\frac{π}{2} π)$ .

f

f (x) = log cos x

(0, \frac{π}{2})

(\frac{π}{2}, π)

Prove that the function f given by f(x) = log sin x is strictly increasing on and strictly decreasing on

Prove that the function f given by f(x) = log cos x is strictly decreasing on and strictly increasing on

Join BYJU'S Learning Program