HCH1: Geometrical Applications of Differentiation

Question 1

The gradient function

Answer 1

y = f’(x)

Question 2

The sign of the derivative

Answer 2

if f’(x) > 0, the function f(x) is increasing

if f’(x)

Question 3

Stationary points

Answer 3

f’(x) = 0

Question 4

Turning points

Answer 4

• Maximum turning points
  f ‘ (x) = 0
  f ‘ (x) > 0 immediately before the point
  f ‘ (x) 0 immediately after the point

Question 5

Absolute maxima and minima

Answer 5

• Greatest value of the function
• Occur either at a maximum/minimum turning point or an endpoint of the domain
• Found by using an inequality provided and subbing values in

Question 6

The second derivative

Answer 6

• Derivative of first derivative

- notations: f’‘(x), y’’, d2y/dx2

Question 7

Points of inflection

Answer 7

• Any point where curve changes concavity
• y’’ = 0 and y’’ for any value before/after is different
• Horizontal POI if y’ = 0

Question 8

Using the second derivative to find the nature of turning points

Answer 8

• If y’’ > 0 at the stationary point, curve is concave up, and turning point is minimum
• If y’’

Question 9

When both the first and second derivatives = 0

Answer 9

• If the curve changes concavity at the SP it will be a horizontal point of inflection
• If the curve is concave up immediately before and after, it will be a minimum tp
• If the curve is concave down immediately before and after, it will be a maximum tp

Question 10

Steps for curve sketching

Answer 10

1. Find where curve cuts axes
2. Find stationary points (using y’ = 0) and determine their nature (using y’’ and values obtained for x value of sp)
3. Find any POI (using y’’ = 0) and determine their nature (using y’’ and x value obtained for x value of POI)
4. Find any symmetry properties (find halfway between x points)
5. Consider behaviour of curve for very large values of x (+and -)
6. Consider set of values for which curve is defined and if there are any asymptotes

Question 11

Equations of tangents to curves

Answer 11

• Use first derivative to find gradient

- Use y-y1 = m(x-x1) equation to solve

Question 12

Equations of normals

Answer 12

• Use first derivative to find gradient of tangent
• Use m1m2 = -1
• Use y-y1 = m(x-x1) equation to solve

Question 13

Primitive functions

Answer 13

• Opposite of differentiation

- raise power of value and then divide by new value, + C to the end