| (i) |
|
f(x) = 6x - 9 |
|
(ii) |
|
f(x) = sin x |
| (iii) |
|
f(x) = x3 + 2x2 - 1 |
|
(iv) |
|
f(x) = cos x |
Solutions
i) f(x) = 6x - 9
We have to find
= ? ---------------------------------- (i)
f(a + h) = ?
as f(x) = 6x - 9
f(a + h) = 6(a + h) - 9
f(a + h) = 6a + 6h - 9
f(a) = ?
as f(x) = 6x - 9
f(a) = 6(a) - 9
f(a) = 6a - 9
Put the value of f(a + h) and f(a) in the equation (i)
| = |
| (6a + 6h - 9) - (6a - 9) |
| h |
|
ii) f(x) = sin x
We have to find
= ? ---------------------------------- (i)
f(a + h) = ?
as f(x) = sin x
f(a + h) = sin(a + h)
f(a) = ?
as f(x) = sin x
f(a) = sin a
Put the value of f(a + h) and f(a) in the equation (i)
| = |
|
[ |
|
] |
|
∴ [sin x - sin y = 2 sin( (x - y)/2 ) cos( (x + y)/2 )] |
iii) f(x) = x3 + 2x2 - 1
We have to find
= ? ---------------------------------- (i)
f(a + h) = ?
as f(x) = x3 + 2x2 - 1
f(a + h) = (a + h)3 + 2(a + h)2 - 1
f(a) = ?
as f(x) = x3 + 2x2 - 1
f(a) = a3 + 2a2 - 1
Put the value of f(a + h) and f(a) in the equation (i)
| = |
| [(a + h)3 + 2(a + h)2 - 1] - (a3 + 2a2 - 1) |
| h |
|
| = |
| [(a + h)3 + 2(a + h)2 - 1] - a3 - 2a2 + 1 |
| h |
|
| = |
| [a3 + h3 + 3ah(a + h) + 2(a2 + h2 + 2ah) - 1] - a3 - 2a2 + 1 |
| h |
|
| = |
| a3 + h3 + 3a2h + 3ah2 + 2a2 + 2h2 + 4ah - 1 - a3 - 2a2 + 1 |
| h |
|
| = |
| h3 + 3a2h + 3ah2 + 2h2 + 4ah |
| h |
|
| = |
|
[ |
h2 + 3a2 + 3ah + 2h + 4a |
] |
| = |
|
h2 + 3a2 + 3ah + 2h + 4a |
| = |
|
h2 + 3ah + 2h + 3a2 + 4a |
| = |
|
h2 + (3a + 2)h + 3a2 + 4a |
iv) f(x) = cos x
We have to find
= ? ---------------------------------- (i)
f(a + h) = ?
as f(x) = cos x
f(a + h) = cos(a + h)
f(a) = ?
as f(x) = cos x
f(a) = cos a
Put the value of f(a + h) and f(a) in the equation (i)
| = |
|
[ |
|
] |
|
∴ [cos x - cos y = -2 sin( (x - y)/2 ) sin( (x + y)/2 )] |
