Everything Maths Grade 12

7.2 CHAPTER 7. DIFFERENTIAL CALCULUS

point x = 2.

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 = 2. It is simpler to substitute x = 2 at the end of the calculation.

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

f (x +h) = 2(x +h)^2 − 5(x +h)

= 2(x^2 + 2xh +h^2 )− 5 x− 5 h

= 2x^2 + 4xh + 2h^2 − 5 x− 5 h

Step 3 : Calculate limit

lim

h→ 0

f (x +h)−f (x)

h

=

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

h

; h�= 0

= lim

h→ 0

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

h

= lim

h→ 0

4 xh + 2h^2 − 5 h

h

= lim

h→ 0

h(4x + 2h− 5)

h

= lim

h→ 0

4 x + 2h− 5

= 4x− 5

Step 4 : Calculate gradient at x = 2

4 x− 5 = 4(2)− 5 = 3

Step 5 : Write the final answer

The gradient of the tangent to the curve f (x) = 2x^2 − 5 x at x = 2 is 3. This is

also the gradient of thecurve at x = 2.

Example 6: Limits

QUESTION

For the function f (x) = 5x^2 − 4 x + 1, determine the gradient of the tangent to curveat the