TRY YOURSELF - IV

Q 1. Show that the square roots of two successive natural numbers greater than N ² differ by less than ¹ 2N . Q 2. If 2a+ +3b 6c=0, prove that the equation ax²+ + =bx c 0 has at least one real root in (0, 1).

Q 3. Prove that

ln ,

b a b b a

b a a

− < < − where 0< <a b

Q 4. Prove that

1 1

2 1 2 1

tan⁻ x −tan⁻ x < −x x, where x₂ >x₁

* * * * * * * * * * * * *

In the unit on functions, we discussed graphs in great detail but most of the discussion was based on obtaining the given graph by some transformation of one of the standard functions that we encountered previously in the same

chapter. As an example ,

( )

^{2 1 1 2 3}

Our purpose in this section is to discuss more advanced graphs by analyzing their nature using the knowledge of derivatives that we now possess.

Draw the graphs of the following functions:

(a)

( )

2

Solution: In all the questions above, we will evaluate the limits of the functions at various important points within their respective domains which will give us a good idea of the overall behaviour of the particular function being analysed.

Also, notice that f(x) is an odd function.

Now,

( )

1

This information is sufficient to accurately plot the graph

⇒ f x is always increasing; since f'(x) also become 0, it might appear that f(x) is not strictly increasing. However, notice that the set of points where f'(x) becomes 0 will be countable so, according to the reason stated earlier f(x) will be strictly increasing. We now proceed to evaluate the set of points where f'(x) = 0 understand since f '(x) is always non-negative i.e, it does not change sign at these points so that f ''(x) can neither be positive or negative; in other words these points can neither be local maxima nor local minimum; they are inflexion points)

This should be clear from the graph:

(c) ^{f x}

( )

^{= −x3 6x2+11x−6}

( ) ( ) ( )

lim ; lim ; 0 6

x f x x f x f

→+∞ = +∞ →−∞ = −∞ = − .

Also, f(x) can be factorized as ^{f x}

( ) (

^{= −x 1}

)(

^x−2

)(

^x−3

)

so that f(x) has three roots, namely

For more accuracy in graph plotting, 1 2

3

f ±  can also be numerically evaluated.

Based on all this information, the graph has been plotted below.

Fig - 36

x e concavity of the graph changes.

You are urged to find that point by evaluating the sign of f"(x) for different values in (0, )∞

Therefore, near x = 1 and x = 2, f (x) will have an unbounded increase in magnitude. (We will soon see that the lines x = 1 and x = 2 would be called asymptotes to the given curve.)

Also, ^lim

( )

^{0 ;}

( )

^{0 1}

( )

³

{ }

Now,

( ) ( ) ^{( )}

A straight line is called an asymptote to the curve y = f (x) if , in layman’s term, the curve touches the line at infinity (this is not technically correct; we should say that the curve tends to touch the line as infinity is approached or as

x→ ∞)

More accurately, an asymptote to a curve is a line such that the distance from a variable point M on the curve to the straight line approaches zero as the point M recedes to infinity along some branch of the curve.

Referring to the previous example, we see that y = 0 is a horizontal asymptote to

( )

2; Of course, there can be inclined asymptotes also. We now formally distinguish between the three kinds of asymptotes and outline the approach to determine them.

(a) Horizontal asymptotes : If ^lim

( )

This discussion will become more clear with an example.

Let ^{f x}

( )

^{x 1}

The graph is sketched below. The extremum points are x= ±1:

2

Before closing this section with a few more examples, have are a few general steps* to be followed whenever one is encountered with the task of sketching the graph of an arbitrary function f (x):

(i) Find the domain of the given function.

(ii) Determine more of its characteristics; for example, is the function even or odd or neither? Is it periodic? If yes, what is the period? And so on.

(iii) Test the function for continuity and differentiability.

(iv) Find the asymptotes of the graph, if any.

(v) Find the extremum/ inflexion points and the intervals of monotonicity.

(vi) To improve accuracy of the plot, one can always evaluate f (x) at additional points.

This is a very general sequence and mostly the graph would be able to be plotted without necessarily following all the steps. For all our current purposes, these steps are more than sufficient.

Plot the graph of ^{f x}

( )

^{=xe1/ x}

Solution: As far as possible, we will try to stick to the general sequence mentioned above for analysing any given function.

* Quoted from I.A. Maron: Problems in Calculus

Example – 37

*

( )

^1/ 2 ^1/

' ^x 1 ^x

f x xe e

x

− 

= ⋅ + ^{e1/x 1 1}

x

 

=  − 

( ) ( ) ( )

' 0 for , 0 1,

f x > x∈ −∞ ∪ ∞

( ) ( )

' 0 for 0,1

f x < x∈

( )

' 0 for 1;

f x = x= a local minimum point; f(1) = e

e

0

Fig - 41 1

y = x + 1 y

x

1

Sketch the graph of ^{f x}

( )

^{= −x6 3x4+3x2−5.}

Solution: * The domain is obviously

* f(x) is an even function

* Since f(x) is a polynomial function, it is continuous and differentiable on _».

* It is obvious that there are no asymptotes to f(x)

* ^{f '}

( )

^{x =6x5−12x3+6x}

^{=6x x}

(

^{4 −2x2 +1}

)

^{=6x x}

(

²⁻¹

)

²

Example – 38

( )

Therefore, at these four points the convexity of the graph changes:

( ) ( )

^{1 1}

This information is sufficient to accurately draw the graph of the given function.

x = -1, 1 are

Plot the graph of

( )

^{1sin 2 cos .}

f x =2 x+ x Solution: * The domain of f(x) is _».

* ^{f x}

( )

is periodic with period 2π and therefore we need to analyze it only in [0, 2 ]π

* ^{f x}

( )

is continuous and differentiable on

* There are no asymptotes to f(x)

* ^{f '}

( )

^{x =cos 2x−sinx}

We see that f(x) will change its convexity at four different points.

( )

^{11 3 11}

" 0 , sin , 2 sin

2 4 2 4

f x x ^π π ^{−  } π π ^{− }

⇒ > ∀ ∈ +   ∪ −  so that f(x) is concave upwards in these intervals

so that f(x) is concave downwards in these intervals.

*

( )

^{0 1, 0,}

( )

^{2 1}

f = f ^{ }  π2 = f π = .

The graph has been plotted below for [0, 2 ]π

Plot the graph of ^{y= +x ln}

(

^x2−1

)

^.

Solution: * The Domain is given by

2 1 0

x − >

\ [ 1, 1]

⇒ =D » −

* ^{f x}

( )

is continuous and differentiable on D

* 1 1

lim ; lim

x x

y y

+ −

→ = −∞ →− = −∞

1

⇒ = ±x are vertical asymptotes to the curve.

Verify that the graph has no other asymptotes Example – 40

* ^{' 1} ₂² 1 y x

= + x

− ' 0

y = when _x²_{+2x− =1 0}

1 2

⇒ = − ±x

1 2

x= − + does not belong to D;

1 2

x= − − is an extremum point

( )

( )

2

2 2

2 1

" 0

1

y x x

x

⇒ = − + < ∀

−

⇒The curve is always concave downwards so that x= − −1 2 is a point of local maximum.

* _x^lim_→+∞^{y= ∞ ; lim}_x→−∞^{y= −∞}

Based on this data, the graph can be plotted as shown below:

Fig - 44 y

-1- 2 x

-1 0 1

In document Application of Derivatives 13.08.07 (Page 56-71)