Everything Maths Grade 12

CHAPTER 7. DIFFERENTIAL CALCULUS 7.2

point x = a.

SOLUTION

Step 1 : Calculating the gradient at a point

We know that the gradient at a point x is given by:

lim

h→ 0

f (x +h)−f (x)

h

In our case x = a. It is simpler to substitute x = a at the end of the calculation.

Step 2 : Write f (x +h) and simplify

f (x +h) = 5(x +h)^2 − 4(x +h) + 1

= 5(x^2 + 2xh +h^2 )− 4 x− 4 h + 1

= 5x^2 + 10xh + 5h^2 − 4 x− 4 h + 1

Step 3 : Calculate limit

lim

h→ 0

f (x +h)−f (x)

h

=

5 x^2 + 10xh + 5h^2 − 4 x− 4 h + 1− (5x^2 − 4 x + 1)

h

= lim

h→ 0

5 x^2 + 10xh + 5h^2 − 4 x− 4 h + 1− 5 x^2 + 4x− 1

h

= lim

h→ 0

10 xh + 5h^2 − 4 h

h

= lim

h→ 0

h(10x + 5h− 4)

h

= lim

h→ 0

10 x + 5h− 4

= 10x− 4

Step 4 : Calculate gradient at x = a

10 x− 4 = 10a− 4

Step 5 : Write the final answer

The gradient of the tangent to the curve f (x) = 5x^2 − 4 x + 1 at x = 1 is 10 a− 4.

Exercise 7 - 1

Determine the following

1.

lim

x→ 3

x^2 − 9

x + 3

2.

lim

x→ 3

x + 3

x^2 + 3x