Question

The function $ {x^x} $ is increasing, when
(A) $ x > \dfrac{1}{e} $
(B) $ x < \dfrac{1}{e} $
(C) $ x < 0 $
(D) for all $ x $

Answer 1

Hint: In simple terms, increasing function is the function which increases with each increasing value of $ x $ . If the function is always increasing, then the slope of the tangent drawn on the graph of it at any point should always be positive. We can represent slope in terms of derivatives. Use this to solve the above question.

Complete step-by-step answer:
A function $ f $ is called an increasing function in the interval $ (x,y) $ if
 $ x < y \Rightarrow f(x) < f(y) $
In terms of differentiation, we can say that, a function $ f $ is called an increasing function in the interval $ (a,b) $ if
 $ f(x) \geqslant 0 $ $ \forall x \in (a,b) $
We would solve this question using the definition of differentiation
Let $ f(x) = y = {x^x} $
By taking log to both the sides, we can write
 $ \log y = \log {x^x} $
Since, we know that, $ \log {m^n} = n\log m $
We can write the above equation as
 $ \log y = x\log x $
Now, by differentiating both the sides with respect to $ x $ we get
 $ \dfrac{d}{{dx}}\log y = \dfrac{d}{{dx}}(x\log x) $
Now, using the formula,
\[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\], $ \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}} $
And, product rule
 $ \dfrac{d}{{dx}}(uv) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u $
And by using the derivative of composite function
 $ \dfrac{d}{{dx}}f(g(x)) = f'(g(x)).\dfrac{d}{{dx}}g(x) $
We can differentiate the above equation as
 $ \dfrac{1}{y}\dfrac{{dy}}{{dx}} = x\dfrac{d}{{dx}}\log x + \log x\dfrac{d}{{dx}}x $
 $ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = x\dfrac{1}{x} + \log x $
By cancelling the common terms and cross multiplying, we get
 $ \dfrac{{dy}}{{dx}} = y(1 + \log x) $
By substituting the value of $ y $ in it, we get
 $ \dfrac{{dy}}{{dx}} = {x^x}(1 + \log x) $
Now, the function is differentiable if
 $ f'(x) = \dfrac{{dy}}{{dx}} \geqslant 0 $
Which can be written as
 $ {x^x}(1 + \log x) \geqslant 0 $
 $ \Rightarrow 1 + \log x \geqslant 0 $
Rearranging it we can write
 $ \Rightarrow \log x \geqslant - 1 $
By using the property, $ \log y = x \Rightarrow y = {e^x} $
We can write
 $ x \geqslant {e^{ - 1}} $
 $ \Rightarrow x \geqslant \dfrac{1}{e} $
Therefore, from the above explanation, the correct answer is, option (A) $ x > \dfrac{1}{e} $
So, the correct answer is “Option A”.

Note: Increasing function has a slope greater than or equal to zero. And in terms of derivatives, slope is the value of the derivative at the point of tangency. In this question, the key point was to know how to express increasing function mathematically. Knowing the properties of differentiation, log and exponents was important too. Such types of questions need more than one concept to solve them. You need a clear idea of basics and mathematical terms.