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APPLICATION OF DERIVATIVES
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Illustration:
Illustration: Let f(x) = x(x - 1)2, find point at which f(x) have maxima & minima.
Ans: f(x) = x(x - 1)2

f'(x) = 2x(x - 1) + (x - 1)2

f'(x) = (x - 1)(3x - 1)

f'(x) = 0

(x - 1)(3x - 1) = 0

Critical points are x = 1, 1/3

f''(x) = 6x - 4

f''(1) = 6 - 4 = 2 > 0

f''(-1/3) = - 2 < 0

f''(1) > 0 so, at n = 1 it has minima

f''(1/3) < 0 so, maxima at x = 1/3.

Global Maxima & Minima:
(i) Global maxima/minima in [a, b] is a greatest/least value of f(x) in [a, b]

Global maxima/minima in [a, b] would always occur at critical points of f(x) with in [a,b] or end points of interval.
To find global maxima/minima of f(x) in [a, b] find out all critical points of f(x) in [a, b] (i.e. all points at which f'(x) = 0)

Let c1, c2, .......... cn be critical points & f(c1), f(c2), .............. f(cn) be values of function of these points.

Let M1 Global Maxima

M2 Global Minima

Then M1 = max.{f(a), f(c1), f(c2), ............... f(cn), f(b)}

&    M2 = min.{f(a), f(c1), f(c2), ................ f(cn), f(b)}

Illustration: If f(x) = 2x3 - 9x2 + 12x + 6. Discuss global maxima and minima of f(x) in (1, 3).
Ans: f(x) = 2x3 - 9x2 + 12x + 6

f'(x) = 6x2 - 18x + 12

f'(x) = 6(x - 1)(x - 2)

f'(x) = 0   x = 1, 2

f(1) = 11 & 1 - (2) = 10

Since open interval is (1, 3)

Clearly x = 2 is only point in (1, 3) & 1 - (2) = 10

Now,

Dumb Question: How = 11.
Ans: f(x) = 2x2 - 9x2 + 12x + 6

= f(1 + h)

= 2(1 + h)3 - 9(1 + h)2 + 12(1 + h) + 6

= 2(1 + h3 + 3h2 + 3h) - 9(1 + h2 + 2h) + 12 + 12h + 6

= 11

So, x = 2 is point of global minima in (1, 3) & global maxima doesnot exist in (1, 3) .

Minima of discontinuous function:
For minima at x = a, 4 cases arises.

From figure f(CD) < f(a + h)           From figure f(a) < f(a + h)
f(a) < f(a - h)                               f(a)

f(a - h)

From figure, f(a) < f(a + h)           From figure, f(a)

f(a + h)
f(a) < f(a - h)                                f(a) < f(a - h)
From all above case, for minima of discontinuous functions,
f(a)

f(a + h)
2f(a)

f(a - h)
Illustration
: Discuss minima of f(x) = {x}, (where {} is raction part of x) for x = 6.

Ans: For discont functions, maximum & minimum at x = a is attained when
f(a)

f(a + h) & f(a)

f(a - h)
Now, f(x) = {x} is discontinuous function at x = 6
Since f(c) = {6} = 0
f(6 + h) = {6 + h} = h > 0
&   f(6 - h) = {6 - h} = 1 - h > 0
So, f(6) < f(6 + h) & f(6) < f(6 - h)

f(x) is minimum at x = 6.
Dumb Question
: How f(6) = {6} = 0

Ans: {x} is fractional function. Since6 is integer & no fractional part

f(6) = 0
Maxima of discontinuous function
:

From figure, f(a) > f(a + h)             From figure, f(a) > f(a + h)
f(a) > f(a - h)                                f(a) > f(a - h)

From figure, f(a)

f(a - h)              From figure, f(a)

f(a + h)
f(a) > f(a + h)                               f(a) > f(a - h)
Prove all cases, maxima of discontinuous function,
f(a)

f(a + h)   &   f(a)

f(a - h)
Illustration
: f(x) =

, then for f(x) at x = 1 discuss maxima & minima.

Ans:

f(x) =

f(1) = 6

f(1 - h) = 6
&  f(1 + h) = 7 - (1 + h) = 6 - h < 6
So, at x = 1 is neither point of maxima nor manima.
Easy Type
:

Q.1. If s = t
3
- 4t, find acceleration at time when velocity is zero.

Ans: s = t
3
- 4t

v =

= 3t
2
- 4 ................................................. (i)
a =

= 6t ....................................................... (ii)
time at which velocity is zero
3t
2
- 4 = 0

t
2
=

a =

Q.2. If r be radius, s the surface atrea & v the volume of spherical buble, prove that
(i)

(ii)

Since v =

r
3

(i)

.................................. (i)

(ii) s = 4

r
2

Q.3. On curve x
3
= 12y, find interval at which abscissa changes at a faster rate than ordinate.

Ans: x
3
= 12y
differencing w.r.t. y

Since abscissa change faster than ordinate

> 1   or

> 1

> 0 where x

0

x
2
- 4 < 0

(x - 2)(x + 2) < 0

- 2 < x < 2

{0}
So, x

(- 2, 2) - {0} is required interval.
Q.4. Find eq. of tangent to parabola y
2
= 4ax at point (at
2
, 2at)

Ans: y
2
= 4ax ............................................................ (i)
differentiating (i) w.r.t. x,

Eq. of tangent at (at
2
, 2at) is
(y - 2at) =

y - 2at =

(x - at
2
)

 yt = x + at2

Q.5. Find a cute angle b/w curves y = |x
2
- 1| & y = |x
2
- 3| at their points of intersection when x > 0.

Ans: For intersection of given curves
|x
2
- 1| = |x
2
- 3|

(x
2
- 1)
2
= (x
2
- 3)
2

2x
2
= 4

x = ±

Since   x > 0, so, x =

is only point of intersection
y = |x
2
- 1| = (x
2
- 1) since x =

y = |x
2
- 3| = - (x
2
- 3) since x =
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