Stationary Points and Gradients in Calculus

Teachings for Teachings for Exercise 12I Exercise 12I

Differentiation You can determine the nature of stationary points by considering gradients. y A point where f(x) stops increasing and starts decreasing is called a maximum point Local maximum A point where f(x) stops decreasing and starts increasing is called a minimum point x Point of inflexion A point of inflexion is where the gradient is locally a maximum or minimum (the gradient does not have to change from positive to negative, for example) Local minimum These are all known as turning points, and occur where f (x) = 0 12I

Differentiation You can determine the nature of stationary points by considering gradients. y To find the coordinates of these points, you need to: Local maximum 1) Differentiate f(x) to get the Gradient Function x Point of inflexion 2) Solve f (x) by setting it equal to 0 (as this represents the gradient being 0) Local minimum 3) Substitute the value(s) of x into the original equation to find the corresponding y-coordinate 12I

Differentiation The second derivative will tell us if the stationary point is a maximum or minimum y Notice how the diagram's gradient is always getting smaller x It starts very positive (and steep), gets less positive, equals 0, then becomes slightly negative, before becoming more negative The gradient's gradient is negative We describe this as ? ?? ?? < 0 ?? When ?2? ??2< 0 Or ?2? ??2< 0 we have a local maximum. 12I

Differentiation The second derivative will tell us if the stationary point is a maximum or minimum y Notice how the diagram's gradient is always getting bigger x It starts very negative (and steep), gets less negative, equals 0, then becomes slightly positive, before becoming more positive The gradient's gradient is positive We describe this as ? ?? ?? > 0 ?? When ?2? ??2> 0 Or ?2? ??2> 0 we have a local minimum. 12I

Differentiation All three points happen when ?? ??= 0 y ?2? ??2< 0 Local maximum x Point of inflexion ?2? ??2= 0 This does not mean that whenever ?2? Local minimum ??2= 0 ?2? ??2> 0 there is a point of inflexion, it could also be a local max or min, but every point of inflexion has ?2? If point of inflexion ?2? ??2= 0 ??2= 0. But A little more work is required here, but I will discuss this later If ?2? ??2= 0 Point of inflexion (it does suggest we should look into it though) 12I

Differentiation y Local maximum x Point of inflexion Local minimum ?? ??> 0 ?? ??< 0 ?? ??> 0 Apart from at point of inflexion where ?? ??= 0 12I

Differentiation Example Question You can determine the nature of stationary points by considering gradients. Find the coordinates of the turning point on the curve y = x4 32x, and state whether it is a minimum or maximum. To find the coordinates of these points, you need to: = 4 32 y x x Differentiate dy dx= 4 x 1) Differentiate f(x) to get the Gradient Function 3 4 32 x Set equal to 0 = 3 32 x = x = 0 2) Solve f (x) by setting it equal to 0 (as this represents the gradient being 0) Add 32 3 4 32 2 Divide by 4, then cube root 3) Substitute the value(s) of x into the original equation to find the corresponding y-coordinate = 4 4 32 y y = y = x (2) 48 x Sub 2 into the original equation Work out the y- coordinate 32(2) The stationary point is at (2, -48) 12I

Differentiation Example Question You can determine the nature of stationary points by considering gradients. Find the coordinates of the turning point on the curve y = x4 32x, and state whether it is a minimum or maximum. To find the coordinates of these points, you need to: = 4 32 y x x The stationary point is at (2, -48) dy dx= 1) Differentiate f(x) to get the Gradient Function 3 4 32 x Differentiate again 2 d y dx 2) Solve f (x) by setting it equal to 0 (as this represents the gradient being 0) = 2 12 x So the stationary point is a MINIMUM in this case! 2 2 12x 3) Substitute the value(s) of x into the original equation to find the corresponding y-coordinate Sub in the x coordinate 2 12(2) Positive = Minimum Negative = Maximum The rate of change of the rate of change is positive This means the rate of change (gradient) is about to increase! = 48 12I

Differentiation Example Question You can determine the nature of stationary points by considering gradients. Find the stationary points on the curve: y = 2x3 15x2 + 24x + 6, and state whether they are minima, maxima or points of inflexion To find the coordinates of these points, you need to: = + + 3 2 2 15 24 6 y x x x Differentiate + = 2 30x 24 '( ) x 6 f x 1) Differentiate f(x) to get the Gradient Function Set equal to 0 + = 2 6 30 24 0 x x Factorise 2) Solve f (x) by setting it equal to 0 (as this represents the gradient being 0) + = 2 6( 5 4) 0 x x Factorise again 4 = 6( x = 4)( OR 1) x = 0 x x Write the solutions 1 3) Substitute the value(s) of x into the original equation to find the corresponding y-coordinate Substituting into the original formula will give the following coordinates as stationary points: (1, 17) and (4, -10) 12I

Differentiation Stationary points at: (1, 17) and (4, -10) Example Question You can determine the nature of stationary points by considering gradients. Find the stationary points on the curve: y = 2x3 15x2 + 24x + 6, and state whether they are minima, maxima or points of inflexion To find the coordinates of these points, you need to: = + + 3 2 2 15 24 6 y x x x + = 2 30x 24 '( ) x 6 f x 1) Differentiate f(x) to get the Gradient Function Differentiate again = ''( ) 12 x 30 f x 2) Solve f (x) by setting it equal to 0 (as this represents the gradient being 0) Sub in x = 1 Sub in x = 4 = = ''( ) 12 x ''(1) 12(1) 30 = ''(1) 18 = 30 ''( ) 12 x ''(4) 12(4) 30 = ''(4) 18 = 30 f f f x f f f x 3) Substitute the value(s) of x into the original equation to find the corresponding y-coordinate So (1,17) is a Maximum So (4,-10) is a Minimum 12I

Differentiation You can determine the nature of stationary points by considering gradients. Please note that if ? ? = 0, the stationary point could be a minimum, maximum or a point of inflexion. In this case, find the gradient just before and just after the coordinate, this will then tell you the nature of the point. 12I