Grade 12 Chapter 6 Application of derivatives

For a quantity y varying with another quantity x, satisfying the rule y = f(x), the rate of change of y with respect to x is given by
dy dx or f ( x0 ) .
x x0

The rate of change of y with respect to the point x = x0 is given by

dy or f ( x) . dx

If the variables x and y are expressed in form of x = f(t) and y = g(t), then the rate of change of y with respect to x is given by
dy dx g (t ) , provided f (t ) 0 f (t )

A function f: (a, b) R is said to be increasing on (a, b), if x1 < x2 in (a, b) f ( x1 ) f ( x2 ) x1 , x2 (a, b) decreasing on (a, b), if x1 < x2 in (a, b) f ( x1 ) f ( x2 ) x1 , x2 (a, b) OR If a function f is continuous on [a, b] and differentiable on (a, b), then f is increasing in [a, b], if f ( x) 0 for each x (a, b) f is decreasing in [a, b], if f ( x) 0 for each x (a, b) f is constant function in [a, b], if f ( x) 0 for each x (a, b) A function f: (a, b) R is said to be strictly increasing on (a, b), if x1 < x2 in (a, b) (a, b) strictly decreasing on (a, b), if x1 < x2 in (a, b) (a, b)

f(x1) < f(x2) f(x1) > f(x2)

x1, x2 x1, x2

The graphs of various types of functions can be shown as follows:

Example 1: Find the intervals in which the function f given by f ( x) 3 sin x cos x, x [0, 2 ] is strictly increasing or decreasing. Solution:
f ( x) f ( x) 3 sin x cos x 3 cos x sin x

f ( x) 0 gives tan x 3 2 5 x , 3 3 2 5 The points x and x divide the interval [0, 2] into three disjoint 3 3 2 2 5 5 intervals, 0, , , , , 2 . 3 3 3 3 2 5 Now, f ( x) 0, if x 0, , 2 3 3 2 5 f is strictly increasing in the intervals 0, and , 2 . 3 3

Also, f ( x) 0, if x

2 5 , 3 3 2 5 . , 3 3

f is strictly decreasing in the interval

For the curve y = f(x), the slope of tangent at the point (x0, y0) is given by
dy dx or f ( x0 ) .
x0 , y0

For the curve y = f(x), the slope of normal at the point (x0, y0) is given by
1 dy dx
x0 , y0

or

1 . f ( x0 )

The equation of tangent to the curve y = f(x) at the point (x0, y0) is given by,
y y0 f ( x0 x x0 )

If f ( x0 ) does not exist, then the tangent to the curve y = f(x) at the point (x0, y0) is parallel to the y-axis and its equation is given by x = x0 The equation of normal to the curve y = f(x) at the point (x0, y0) is given by,
y y0 1 ( x x0 ) f ( x0 )

If f ( x0 ) does not exist, then the normal to the curve y = f(x) at the point (x0, y0) is parallel to the x-axis and its equation is given by y = y0 If f ( x0 ) = 0, then the respective equations of the tangent and normal to the curve y = f(x) at the point (x0, y0) are y = y0 and x = x0 Let y = f(x) and let x be a small increment in x and y be the increment in y corresponding to the increment in x i.e., y = f(x + x) f(x) Then, dy
f ( x)dx or dy dy dx x is a good approximation of y, when dx =

x is relatively small and we denote it by dy

Maxima and Minima: Let a function f be defined on an interval I. Then, f is said to have maximum value in I, if there exists c I such that f(c) > f(x), x I [In this case, c is called the point of maxima] minimum value in I, if there exists c I such that f(c) < f(x), x I [In this case, c is called the point of minima]

an extreme value in I, if there exists c I such that c is either point of maxima or point of minima [In this case, c is called an extreme point] Note: Every continuous function on a closed interval has a maximum and a minimum value. Local maxima and local minima: Let f be a real-valued function and c be an interior point in the domain of f. Then c is called a point of local maxima, if there exists h > 0 such that f(c) > f(x), x (c h, c + h) [In this case, f(c) is called the local maximum value of f] local minima, if there exists h > 0 such that f(c) < f(x), x (c h, c + h) [In this case, f(c) is called the local maximum value of f]

A point c in the domain of a function f at which either f (c) 0 or f is not differentiable is called a critical point of f. First derivative test: Let f be a function defined on an open interval I. Let f be continuous at a critical point c in I. Then: If f ( x) changes sign from positive to negative as x increases through c, i.e. if f ( x) 0 at every point sufficiently close to and to the left of c, and f ( x) 0 at every point sufficiently close to and to the right of c, then c is a point of local maxima. If f ( x) changes sign from negative to positive as x increases through c, i.e. if f ( x) 0 at every point sufficiently close to and to the left of c, and f ( x) 0 at every point sufficiently close to and to the right of c, then c is a point of local minima. If f ( x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. Such a point c is called point of inflection. Second derivative test: Let f be a function defined on an open interval I and c I. Let f be twice differentiable at c and f (c) = 0. Then:

If f (c) 0 , then c is a point of local maxima. In this situation, f(c) is local maximum value of f. If f (c) 0 , then c is a point of local minima. In this situation, f(c) is local minimum value of f. If f (c) 0 , then the test fails. In this situation, we follow first derivative test and find whether c is a point of maxima or minima or a point of inflection. Example 2: Find all the points of local maxima or local minima of the function f given by f(x) = x3 12x2 + 36x 4 Solution: We have
f ( x) x 3 12 x 2 36 x 4 f ( x) 3 x 2 24 x 36 3( x 2 8 x 12) and f ( x) 3(2 x 8) 6( x 4)

Now,
f ( x) 0 gives x 2 8 x 12 0 ( x 2)( x 6) x 2 or x 6 12 and f (6) 12 However, f (2) 0

Therefore, the point of local maxima and local minima are at the points x = 2 and x = 6 respectively. The local maximum value is f(2) = 28 The local minimum value is f(6) = 4 Absolute maximum value or absolute minimum value: Let f be a differentiable and continuous function on a closed interval, then f always attains its maximum and minimum value in the interval I, which are respectively known as the absolute maximum and absolute minimum value of f. Also, f attains these values at least once each in [a, b]. Let f be a differentiable function on a closed interval I and c be any interior point of I such that f (c) 0 , then f attains its absolute maximum value and its absolute minimum value at c. To find the absolute maximum value or/and absolute minimum value, we follow the steps listed below: Step 1: Find all critical points f in the interval. Step 2: Take the end point of interval. Step 3: Calculate the values of f at the points found in step 1 and step 2.

Step 4: Identify the maximum and minimum values of f out of values calculated in step 3. The maximum value will be the absolute maximum (greatest) value of f and the minimum value will be the absolute minimum (least) value of f. Example 3: Find the absolute maximum value and/ or minimum absolute value of the function f(x) = 3x4 x2, x [1, 1] Solution: We have,
f ( x) 3 x 4 x2 f ( x) 12 x3 2 x Now, f ( x) 0 gives

2 x(6 x 2 1) 0 x 0 or x 1 1 or 6 6

Now, f(1) = 2 f(1) = 2 f(0) = 0

Clearly, absolute maximum value and absolute minimum value of f in [1, 1] are 2 and
1 respectively. 12

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