Page 4 - Calculus

Gradient of a function at any point.

Example
Determine the gradient of of the following:
3
2
a. ( ) = 2 + 2 + 5 + 7 at = 3
4
b. ( ) = − − 5 + 3 at = −2

Solution
Gradient = f'(x)
2
a. ′( ) = 6 + 4 + 5
2
thus gradient at = 3, ′(3)= 6(3) + 4(3) + 5 = 71
b. ( ) = −4 −1 − 5 + 3
∴ f'(x) = 4 −2 − 5
gradient, at = −2, f'(−2) = 4(−2) −2 − 5 = −4

Average Gradient

Average gradient between = and = of ( ) is given by

= ( ) − ( )
−
-
Example
1. Determine the average gradient of
3
2
a. ( ) = −4 + 6 − 2 + 5 between = −1 and = 3
2
b. ( ) = 3 + 3 − 1 between = 0 and = 4
2
c. ( ) = + 3 between = 2 and = 6
x−4
1
d. ( ) = 2 ( ) − 4 between = −2 and = 0
3
Solutions
3
2
a. (−1 ) = −4(−1) + 6(−1 ) − 2(−1 ) + 5 = 17
2
3
(3 ) = −4(3) + 6(3 ) − 2(3 ) + 5 = −55
m average = ( ) − ( )
−

= 17−(−55)
−1−3
= −18
2
b. (0) = 3(0) + 3(0) − 1 = −1
(4) = 3(4) + 3(4) − 1 = 59
2
m average = ( ) − ( )
−

= −1−59
0−4
= 15