in Calculus retagged ago by
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17 votes
17 votes

Consider the functions

  1.  $e^{-x}$
  2. $x^{2}-\sin x$
  3. $\sqrt{x^{3}+1}$

Which of the above functions is/are increasing everywhere in  $[ 0,1]$?

  1. Ⅲ only
  2. Ⅱ only
  3. Ⅱ and  Ⅲ only
  4. Ⅰ and Ⅲ only
in Calculus retagged ago by
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2 Comments

III) only

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sin1= sin57.3º
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3 Answers

49 votes
49 votes
Best answer

Answer A. III Only   (i.e. $\sqrt{x^3 + 1}$ )

Explanation

Decreasing/Increasing nature of a function can be determined by observing the first derivative of equations in given domain.

If the derivative is positive in given domain, It is increasing, else a negative value indicates it is decreasing.

Now testing one by one

I. $e^{-x}$

Here, $\frac{\mathrm{d} }{\mathrm{d} x}e^{-x} = -e^{-x}$

This will remain negative in entire domain [0,1] hence decreasing.

II. $x^{2} - \sin(x)$

Here, $\frac{\mathrm{d} }{\mathrm{d} x}(x^{2} - \sin(x)) = 2x - \cos(x)$

Here the switch happens! Observe $2x - \cos(x)$ is negative for $x=0$ (i.e. = -1) and positive at $x=1$ (i.e. = ~1.4596977). Though we can find the exact point where it switches but that's not required here. We can confirm that till some point in the domain [0,1], this function decreases and then increases.

III. $\sqrt{x^3 + 1}$

Here, (Though this one is intuitive), $\frac{\mathrm{d} }{\mathrm{d} x}\sqrt{x^3 + 1} = \frac{3x^{2}}{2\sqrt{x^3+1}}$   This will remain positive throughout [0,1] and hence will be increasing.

edited by

3 Comments

After 800+ views, I don't know why nobody has given upvote to this answer which explains the correct procedure.
For, $2^{nd}$ function, we can do like this :
$f(x) = x^2 - sinx \Rightarrow f'(x) = 2x - cosx$
Now, if $f'(x) = 0$ which means $2x - cosx = 0 \Rightarrow 2x = cosx$
Now, RHS $cosx$ lies between $-1$ and $+1$ [including both points] for all real $x$, So, LHS $2x$ should also lies between  $-1$ and $+1$
So, $-1 \leq 2x \leq +1 \Rightarrow -1/2 \leq x \leq +1/2$
So, for the given question,  if  $f'(x) = 0$ then $0 \leq x \leq +1/2$
Same thing is true for other way means if $x$ is between $0$ and $1/2$ then $f'(x)=0$
So, $f'(x) = 0$ iff $x \in [0,1/2]$, Since, $f'(x)$ is not $>0$, So, it is not strictly increasing. Here, also, $f''(x) > 0$ in the given interval because $f''(x) = 2+sinx$. Since $sinx$ lies between $-1$ and $+1$ for all real $x$. So, $f''(x) = 2+sinx$ will lie between $1$ and $3$. Since, $f'(x)=0$ for interval $[0,1/2]$ and $f''(x)>0$, So, we can say $f(x)$ has atleast one local minima in the given interval. Hence, it can't be  neither strictly increasing nor non-decreasing.
--  For the first function, if someone knows  graph of $e^x$ but not $e^{-x}$. To make the graph of $f(-x)$ from $f(x)$, just reflect the $f(x)$ about y-axis means $y-axis$ will work as mirror. So, here reflect $e^x$ about y-axis.
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Nice explanation! Please, Correct your derivation of sqrt(x^3 + 1). It's wrong.
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@rajankakaniya

Corrected! Thanks for pointing this out!

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5 votes
5 votes
A. 3 only .

1. e^-x is popular decreasing function.

2. if you will put 0.5 ( used virtual calculator) you will get some -ve value in x^2 - sinx . at 0 function value is 0 at 0.5 it is -ve it means not increasing function.

3. very easy to tell increasing function.

3 Comments

sin(0.5)=0.008,

0.25-0.008=0.24, positive value...
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@scarz sin (0.5) = 0.47
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But, then finding the decreasing co-ordinate by random assumptions might take extra time. :(
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1 vote
1 vote
III Only.

Increasing fun means:For all values of x within the domain of the fun,

Increasing in x will result increasing in f(x)

&&

Dcreasing in x will result decreasing in f(x).

Now,for option I,

x=0----->f(x)=1;x=1----->f(x)=0.368

[Increasing x does not result increasing f(x)]

For option II,

X=0----->f(x)=0;x=0.5----->f(x)=0.25-sin(180*0.5/3.14)=-0.22

[Increasing x does not result increasing f(x)]

Note that value of x is in radian,so u have converted in degree to use it inside sin function.

For option III,

Clearly f(x) increases/decreases whenever u increase/decrease x within [0,1].