3.4 - Gradients and turning points

3.4 - Gradients and turning points

Differentiation can locate a curve's horizontal tangents and points with any chosen gradient. You will find their coordinates, then use the original graph's shape to identify maxima and minima. A zero gradient does not always mean that the graph turns.

Finding points with a given gradient (Higher)

If a curve has zero gradient, must it have reached its lowest point? Keep that question in mind as you learn to locate gradients first and interpret the shape afterwards.

You already know

Differentiate axnax^n to anxn−1anx^{n-1} for nonzero integer nn; constants give zero. A derivative gives gradient, while the original function gives height. Solve an equation to find an unknown input.

For y=f(x)y=f(x), finding the gradient at a known input aa means evaluating f′(a)f'(a). Finding where the gradient equals a given number mm means solving f′(x)=mf'(x)=m. These operations are different.

A specified gradient

f′(x)=m,y=f(x)f'(x)=m,\qquad y=f(x)

Solve the first equation for every permitted input; use the second for the heights. These differentiation and gradient rules are not in the Higher formulae sheet: learn them. This lesson is Higher only.

For example, y=x2−6x+4y=x^2-6x+4 has derivative 2x−62x-6, so its gradient at x=1x=1 is −4-4. To find zero gradient instead, solve 2x−6=02x-6=0, giving x=3x=3, and then y=9−18+4=−5y=9-18+4=-5. The point is (3,−5)(3,-5), not (3,0)(3,0).

Worked example 1: two points with one gradient

Question. Find all points on y=x3−9x+2y=x^3-9x+2 where the gradient is 33.

Plan. Differentiate, set the derivative equal to 33, then use the original function for heights.

dydx=3x2−9differentiate3x2−9=3require gradient 3x2=4 ⇒ x=−2,2keep both rootsy(−2)=−8+18+2=12original functiony(2)=8−18+2=−8original function\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=3x^2-9 &&\text{differentiate}\\ 3x^2-9&=3 &&\text{require gradient }3\\ x^2&=4\ \Rightarrow\ x=-2,2 &&\text{keep both roots}\\ y(-2)&=-8+18+2=12 &&\text{original function}\\ y(2)&=8-18+2=-8 &&\text{original function} \end{aligned}

Why this step? The derivative identifies the inputs; it cannot supply the graph's heights.

Answer. (−2,12)(-2,12) and (2,−8)(2,-8).

Check. 3(−2)2−9=33(-2)^2-9=3 and 3(2)2−9=33(2)^2-9=3, as required.

Change only the requested gradient. The curve is unchanged, but the required points move.

Worked example 2: a steeper requested gradient

Question. Find all points on y=x3−9x+2y=x^3-9x+2 where the gradient is 1818.

Plan. Solve the new derivative equation, then calculate both heights.

3x2−9=18new gradient targetx2=9 ⇒ x=−3,3keep both signsy(−3)=−27+27+2=2first heighty(3)=27−27+2=2second height\begin{aligned} 3x^2-9&=18 &&\text{new gradient target}\\ x^2&=9\ \Rightarrow\ x=-3,3 &&\text{keep both signs}\\ y(-3)&=-27+27+2=2 &&\text{first height}\\ y(3)&=27-27+2=2 &&\text{second height} \end{aligned}

Why this step? Setting the derivative to zero automatically would answer a different question.

Answer. (−3,2)(-3,2) and (3,2)(3,2).

Check. At either input, 3x2−9=27−9=183x^2-9=27-9=18.

A negative gradient means the curve falls as the input increases. It does not mean that its height is negative. When the variables have units, the derivative's units are output units per input unit.

Watch out

After finding x=2x=2 in the first example, writing (2,3)(2,3) uses the gradient as the height. Substitute into the original curve: y(2)=−8y(2)=-8, so the point is (2,−8)(2,-8).

Find points for another gradient on the same curve.

Finding stationary coordinates (Higher)

A stationary point has a horizontal tangent: the derivative is zero there. Its height can be positive, negative or zero. The derivative tells you where it is stationary; the graph's shape will tell you what kind of stationary point it is.

Solve dy/dx=0\mathrm{d}y/\mathrm{d}x=0, keeping all roots in the domain, and substitute each into the original function. This stationary-point condition is not in the Higher formulae sheet: learn it.

Worked example 3: keeping the zero root

Question. Find all stationary points of y=x3−3x2+1y=x^3-3x^2+1.

Plan. Differentiate and factorise the zero-gradient equation without dividing away a possible root.

dydx=3x2−6xdifferentiate3x(x−2)=0factorise the stationary equationx=0 or 2zero product ruley(0)=1first heighty(2)=8−12+1=−3second height\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=3x^2-6x &&\text{differentiate}\\ 3x(x-2)&=0 &&\text{factorise the stationary equation}\\ x&=0\ \text{or}\ 2 &&\text{zero product rule}\\ y(0)&=1 &&\text{first height}\\ y(2)&=8-12+1=-3 &&\text{second height} \end{aligned}

Why this step? Factorising keeps the solution x=0x=0, which would disappear if you divided by xx.

Answer. (0,1)(0,1) and (2,−3)(2,-3).

Check. Both inputs give derivative zero; neither height has to be zero.

Change only the curve's constant. The stationary inputs stay fixed, while the heights move.

Worked example 4: moving stationary heights

Question. Find all stationary points of y=x3−3x2+5y=x^3-3x^2+5.

Plan. Solve the same derivative equation, then use the new original function.

dydx=3x2−6xconstant gives zero3x(x−2)=0 ⇒ x=0,2stationary inputsy(0)=5new first heighty(2)=8−12+5=1new second height\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=3x^2-6x &&\text{constant gives zero}\\ 3x(x-2)&=0\ \Rightarrow\ x=0,2 &&\text{stationary inputs}\\ y(0)&=5 &&\text{new first height}\\ y(2)&=8-12+5=1 &&\text{new second height} \end{aligned}

Why this step? A vertical translation preserves gradients but changes every graph height.

Answer. (0,5)(0,5) and (2,1)(2,1).

Check. Both points are exactly 44 units above the previous curve's stationary points.

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