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] (i.e. all points at which f’(x) = 0)
Let c₁, c₂, ………. c_n be critical points & f(c₁), f(c₂), ………….. f(c_n) be values of function of these points.
Let M₁ Global Maxima
     M₂ Global Minima
Then M₁ = max.{f(a), f(c₁), f(c₂), …………… f(c_n), f(b)}
  &    M₂ = min.{f(a), f(c₁), f(c₂), ……………. f(c_n), 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 = t³ - 4t, find acceleration at time when velocity is zero.

Ans: s = t³ - 4t
   v = = 3t² - 4 ................................................. (i)
       a = = 6t ....................................................... (ii)
time at which velocity is zero
3t² - 4 = 0   t² =
a =


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

Since v = r³
(i) .................................. (i)

(ii) s = 4r²



Q.3. On curve x³ = 12y, find interval at which abscissa changes at a faster rate than ordinate.

Ans: x³ = 12y
differencing w.r.t. y

Since abscissa change faster than ordinate
> 1   or   > 1
> 0 where x 0
x² - 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 y² = 4ax at point (at², 2at)

Ans: y² = 4ax ............................................................ (i)
differentiating (i) w.r.t. x,

Eq. of tangent at (at², 2at) is
      (y - 2at) =
   y - 2at = (x - at²)

yt = x + at2

Q.5. Find a cute angle b/w curves y = |x² - 1| & y = |x² - 3| at their points of intersection when x > 0.

Ans: For intersection of given curves
    |x² - 1| = |x² - 3|   (x² - 1)² = (x² - 3)²
2x² = 4   x = ±
Since   x > 0, so, x = is only point of intersection
y = |x² - 1| = (x² - 1) since x =
y = |x² - 3| = - (x² - 3) since x =