Stationary Points, Increasing and Decreasing Functions

\mathrm{f}'(x) is the gradient of the graph of y=\mathrm{f}(x): think about which way the graph goes when its gradient is negative.

1. \mathrm{f}'(x) is the gradient of the graph of y=\mathrm{f}(x)
2. A negative gradient means the graph goes down as x increases, so f is decreasing
3. It says nothing about the sign of \mathrm{f}(x) itself
4. \mathrm{f}'(x) is never 0 for 1<x<5, so there is no stationary point (and no maximum) there
5. Answer: f is a decreasing function for 1<x<5

Stationary points are where \dfrac{dy}{dx}=0: differentiate, then solve the quadratic equation you get.

1. \dfrac{dy}{dx}=3x^2-6x-24
2. At a stationary point 3x^2-6x-24=0
3. Divide by 3: x^2-2x-8=0
4. (x-4)(x+2)=0
5. Answer: x=-2 and x=4

At a stationary point \dfrac{dy}{dx}=0

1. \dfrac{dy}{dx}=3x^2+2kx
2. At a stationary point \dfrac{dy}{dx}=0
3. Substitute x=4: 48+8k=0
4. k=-6

Substitute into \dfrac{dy}{dx}, not into y; a gradient going from negative to positive is the shape of a valley.

1. \dfrac{dy}{dx}=x^2-2x-8 (and at x=4 this is 16-8-8=0)
2. At x=3: 9-6-8=-5, negative
3. At x=5: 25-10-8=7, positive
4. Gradient goes negative, zero, positive: the curve goes down then up
5. Answer: a minimum, because the gradient is negative at x=3 and positive at x=5

The curve y=\mathrm{f}(x) has its stationary points where the gradient is zero; look at the sign of the gradient either side of each one.

1. y=\mathrm{f}(x) has stationary points where \dfrac{dy}{dx}=0: at x=-2 and at x=4
2. Just left of x=-2 the graph of \dfrac{dy}{dx} is below the axis (negative gradient)
3. Just right of x=-2 it is above the axis (positive gradient)
4. Negative, zero, positive means a minimum (at x=4 the gradient goes from positive to negative: a maximum)
5. Answer: x=-2

y is increasing where \dfrac{dy}{dx}>0; find where \dfrac{dy}{dx}=0 first.

1. \dfrac{dy}{dx}=3x^2-12x-36=3(x^2-4x-12)
2. =3(x-6)(x+2), which is 0 at x=-2 and x=6
3. \dfrac{dy}{dx} is a positive quadratic, so it is positive outside its roots
4. Answer: x<-2 or x>6

Write \dfrac{8}{x^2} as 8x^{-2}, differentiate twice and substitute x=2

1. y=2x+8x^{-2}
2. \dfrac{dy}{dx}=2-16x^{-3}, which is 0 when x=2
3. \dfrac{d^2y}{dx^2}=48x^{-4}=\dfrac{48}{x^4} (the two minuses make a plus)
4. At x=2: \dfrac{d^2y}{dx^2}=\dfrac{48}{16}=3
5. Positive second derivative, so the point is a minimum.

Picture a cubic with these two turning points and count how many times it meets the x-axis.

1. The maximum is on the left, so the curve goes up, then down, then up again
2. The maximum (-1,\ 0) is on the x-axis, so the curve touches the axis there
3. It goes down to (3,\ -32) and back up, crossing the x-axis once more
4. The others are false: the tangent at a turning point is horizontal (y=-32), the curve is decreasing between the maximum and the minimum, and \dfrac{d^2y}{dx^2}<0 at a maximum
5. Answer: the equation \mathrm{f}(x)=0 has exactly two different solutions

Find both stationary points, then use \dfrac{d^2y}{dx^2} to decide which one is the maximum.

1. \dfrac{dy}{dx}=9-x^2
2. 9-x^2=0 gives x=3 or x=-3
3. \dfrac{d^2y}{dx^2}=-2x
4. At x=3: \dfrac{d^2y}{dx^2}=-6<0, so this is the maximum
5. y=27-\dfrac{27}{3}=27-9=18
6. Answer: (3,\ 18)

Write \dfrac{18}{x} as 18x^{-1}, differentiate, and solve \dfrac{dy}{dx}>0 remembering that x>0.

1. y=2x+18x^{-1}
2. \dfrac{dy}{dx}=2-18x^{-2}=2-\dfrac{18}{x^2}
3. Increasing when 2-\dfrac{18}{x^2}>0
4. Multiply by x^2, which is positive: 2x^2>18, so x^2>9
5. x is positive, so x>3
6. Answer: x>3

Differentiate twice, then factorise \dfrac{d^2y}{dx^2} and decide where the quadratic is below zero.

1. \dfrac{dy}{dx}=4x^3-18x^2+2
2. \dfrac{d^2y}{dx^2}=12x^2-36x
3. 12x^2-36x=12x(x-3), which is 0 at x=0 and x=3
4. A positive quadratic is negative between its roots
5. (So the gradient of the curve is decreasing for these values of x.)
6. Answer: 0<x<3

Write \dfrac{54}{x} as 54x^{-1} before you differentiate.

1. y=x^2+54x^{-1}
2. \dfrac{dy}{dx}=2x-54x^{-2}=2x-\dfrac{54}{x^2}
3. Set \dfrac{dy}{dx}=0: 2x=\dfrac{54}{x^2}, so x^3=27
4. x=3
5. y=9+18=27
6. Answer: (3,\ 27)

Set the gradient equal to 0 to find x, then differentiate again, writing \dfrac{24}{x^2} as 24x^{-2}.

1. At the stationary point 3x-\dfrac{24}{x^2}=0
2. Multiply by x^2: 3x^3=24, so x^3=8 and x=2
3. \dfrac{dy}{dx}=3x-24x^{-2}
4. \dfrac{d^2y}{dx^2}=3+48x^{-3} (the two minuses make a plus)
5. At x=2: 3+\dfrac{48}{8}=3+6
6. Answer: 9 (positive, so the stationary point is a minimum)

Substitute each x-coordinate into \dfrac{dy}{dx}=0 to get two equations in a and b.

1. \dfrac{dy}{dx}=6x^2+2ax+b
2. x=-1: 6-2a+b=0
3. x=4: 96+8a+b=0
4. Subtract the first equation from the second: 90+10a=0, so a=-9
5. b=2a-6=-18-6
6. Answer: b=-24

Differentiate, then take out the coefficient of x^2 before completing the square.

1. \mathrm{f}'(x)=4x^2-12x+10
2. Take out 4 from the x terms: 4(x^2-3x)+10
3. x^2-3x=\left(x-\frac32\right)^2-\frac94
4. So \mathrm{f}'(x)=4\left(x-\frac32\right)^2-9+10
5. (A square is never negative, so \mathrm{f}'(x)\geqslant1 and f is increasing for all x.)
6. Answer: 4\left(x-\frac32\right)^2+1

Differentiate, set \dfrac{dy}{dx}=0 and factorise, then use \dfrac{d^2y}{dx^2} to decide which point is which. Answer parts (a) and (b) before attempting part (c).

Part (a)

1. \dfrac{dy}{dx}=6x^2+6x-36=6(x+3)(x-2)
2. Stationary points at x=-3 and x=2
3. \dfrac{d^2y}{dx^2}=12x+6
4. At x=-3: \dfrac{d^2y}{dx^2}=-30<0, so this is the maximum
5. y=-54+27+108+15=96
6. Answer: (-3,\ 96)

Part (b)

1. At x=2: \dfrac{d^2y}{dx^2}=30>0, so this is the minimum
2. y=16+12-72+15=-29
3. Answer: (2,\ -29)

Part (c)

1. The x^3 term is positive, so the curve rises on the right: not A
2. The stationary points are at x=-3 and x=2: not B
3. The minimum (2,\ -29) is below the x-axis: not D
4. C has a maximum above the axis at x=-3, a minimum below it at x=2 and crosses the y-axis at a positive value (15)
5. Answer: C

Differentiate, set \dfrac{dy}{dx}=0 and take out the common factor 4x first so you don't lose the solution x=0

Part (a)

1. \dfrac{dy}{dx}=4x^3-12x^2-16x
2. Set \dfrac{dy}{dx}=0: 4x(x^2-3x-4)=0
3. 4x(x-4)(x+1)=0
4. x=-1, x=0 or x=4

Part (b)

1. \dfrac{d^2y}{dx^2}=12x^2-24x-16
2. At x=-1: 12+24-16=20>0, minimum
3. At x=0: -16<0, maximum
4. At x=4: 192-96-16=80>0, minimum
5. At x=0, y=5, so the maximum is (0,\,5)

Part (c)

1. The curve has a maximum at x=0 and a minimum at x=4
2. So between them it goes down.
3. 0<x<4

A stationary point gives two equations: the gradient is 0 there, and the point lies on the curve. Answer part (a) before attempting part (b).

Part (a)

1. y=px^2+qx^{-1}, so \dfrac{dy}{dx}=2px-qx^{-2}
2. Stationary at x=2: 4p-\dfrac{q}{4}=0, so q=16p
3. (2,\ 6) is on the curve: 4p+\dfrac{q}{2}=6
4. Substitute q=16p: 4p+8p=6
5. 12p=6
6. Answer: p=\dfrac12

Part (b)

1. q=16p=16\times\dfrac12
2. Answer: q=8

Part (c)

1. \dfrac{dy}{dx}=x-8x^{-2}
2. \dfrac{d^2y}{dx^2}=1+16x^{-3}=1+\dfrac{16}{x^3}
3. At x=2: 1+\dfrac{16}{8}=1+2
4. Answer: 3 (positive, so (2,\ 6) is a minimum)

At a stationary point the tangent is horizontal, so its equation is y= a number. Answer part (a) before attempting part (b).

Part (a)

1. \dfrac{dy}{dx}=6x^2-18x+12=6(x-1)(x-2)
2. Stationary points at x=1 and x=2
3. \dfrac{d^2y}{dx^2}=12x-18
4. At x=2: \dfrac{d^2y}{dx^2}=6>0, so this is the minimum
5. y=16-36+24-1=3
6. Answer: (2,\ 3)

Part (b)

1. The gradient at a minimum point is 0, so the tangent is horizontal
2. It passes through (2,\ 3)
3. Answer: y=3

Part (c)

1. Solve 2x^3-9x^2+12x-1=3
2. 2x^3-9x^2+12x-4=0
3. The tangent touches the curve at x=2, so (x-2)^2 is a factor
4. 2x^3-9x^2+12x-4=(x-2)^2(2x-1)
5. x=\dfrac12, and y=3 on the tangent
6. Answer: Q=\left(\dfrac12,\ 3\right)

f is increasing where \mathrm{f}'(x)>0 and decreasing where \mathrm{f}'(x)<0, so start each part by differentiating.

Part (a)

1. \mathrm{f}'(x)=3x^2-18x+12
2. Decreasing when 3x^2-18x+12<0, that is x^2-6x+4<0
3. Roots: x=\dfrac{6\pm\sqrt{36-16}}{2}=\dfrac{6\pm2\sqrt5}{2}=3\pm\sqrt5
4. A positive quadratic is negative between its roots
5. Answer: 3-\sqrt5<x<3+\sqrt5

Part (b)

1. \mathrm{f}'(x)=3x^2-2ax+12
2. f is increasing for all x when \mathrm{f}'(x)>0 for all x: the quadratic has no real roots
3. Discriminant <0: (-2a)^2-4\times3\times12<0
4. 4a^2<144, so a^2<36
5. -6<a<6, and a is positive
6. Answer: 0<a<6