# application-of-derivatives-6

__ 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 = 0)

Let c

_{1}, c

_{2}, ………. c

_{n}be critical points & f(c

_{1}), f(c

_{2}), ………….. f(c

_{n}) be values of function of these points.

Let M

_{1}Global Maxima

M

_{2}Global Minima

Then M

_{1}= max.{f(a), f(c

_{1}), f(c

_{2}), …………… f(c

_{n}), f(b)}

& M

_{2}= min.{f(a), f(c

_{1}), f(c

_{2}), ……………. f(c

_{n}), f(b)}

__: If f(x) = 2x__

**Illustration**^{3}- 9x

^{2}+ 12x + 6. Discuss global maxima and minima of f(x) in (1, 3).

Ans: f(x) = 2x

^{3}- 9x

^{2}+ 12x + 6

f’(x) = 6x

^{2}- 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,

__: How = 11.__

**Dumb Question**Ans: f(x) = 2x

^{2}- 9x

^{2}+ 12x + 6

= f(1 + h)

= 2(1 + h)

^{3}- 9(1 + h)

^{2}+ 12(1 + h) + 6

= 2(1 + h

^{3}+ 3h

^{2}+ 3h) - 9(1 + h

^{2}+ 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)

__: Discuss minima of f(x) = {x}, (where {} is raction part of x) for x = 6.__

**Illustration**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© = {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.

__: How f(6) = {6} = 0__

**Dumb Question**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)

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

**Illustration**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 = 4r

^{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 + at^{2} |

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 =