G3 - Tangents, normals and monotonicity

G3 - Tangents, normals and monotonicity

Find tangent and normal equations from the point and derivative, including horizontal and vertical cases. Use derivative signs and gradient graphs to justify where a function increases or decreases, with care over isolated zero slopes and gaps in the domain.

Finding a tangent equation

Can the tangent to y=x3y=x^3 cross the curve at the point where it touches? At the origin its gradient is zero, so the tangent is y=0y=0, yet the curve goes from below that line to above it. Tangency is a local match of point and direction; it does not mean the whole line meets the whole curve only once.

You already know

A line through (a,b)(a,b) with finite gradient mm has equation y−b=m(x−a)y-b=m(x-a). Substituting into ff gives a height, while substituting into f′f' gives a gradient. The point-gradient line equation is not supplied in the formula booklet.

Find the point first, differentiate the function and evaluate the derivative at the same input. Keeping the two substitutions separate prevents a height being used as a slope.

Tangent equation

At an input aa where f′(a)f'(a) is finite,

y−f(a)=f′(a)(x−a)y-f(a)=f'(a)(x-a)

Not in the formula booklet: learn it.

The curve's formula supplies both ingredients, but they come from different functions.

Worked example 1: a tangent to a quadratic

Question. Find the tangent to y=x2−3x+1y=x^2-3x+1 at x=2x=2.

Plan. Find the height and derivative separately, then put the point and slope into the line equation.

f(2)=4−6+1=−1point on the curvef′(x)=2x−3differentiatef′(2)=1tangent gradienty+1=1(x−2)point-gradient equation\begin{aligned} f(2)&=4-6+1=-1 && \text{point on the curve}\\ f'(x)&=2x-3 && \text{differentiate}\\ f'(2)&=1 && \text{tangent gradient}\\ y+1&=1(x-2) && \text{point-gradient equation} \end{aligned}

Why this step? The tangent passes through (2,−1)(2,-1), not through (2,1)(2,1); 11 is its gradient, not its height.

Answer. y=x−3y=x-3.

Check. At x=2x=2, the line gives y=−1y=-1, and its coefficient of xx is the required slope 11.

Change only the point's input, from 22 to 11. The curve is unchanged, but its tangent now slopes downwards.

Worked example 2: a falling tangent

Question. Find the tangent to y=x2−3x+1y=x^2-3x+1 at x=1x=1.

Plan. Re-evaluate both the original function and the same derivative at 11.

f(1)=1−3+1=−1heightf′(1)=2(1)−3=−1slopey+1=−(x−1)line through the pointy=−xsimplify\begin{aligned} f(1)&=1-3+1=-1 && \text{height}\\ f'(1)&=2(1)-3=-1 && \text{slope}\\ y+1&=-(x-1) && \text{line through the point}\\ y&=-x && \text{simplify} \end{aligned}

Why this step? Changing the input requires changing both the point and the gradient; neither can be copied from the previous tangent.

Answer. y=−xy=-x.

Check. The curve minus the tangent is x2−2x+1=(x−1)2x^2-2x+1=(x-1)^2, zero at the contact input with no sign change.

AQA accepts equivalent correct line forms when no form is specified. The useful habit is to write the point-gradient equation before simplifying.

Watch out

Wrong: f(2)=−1f(2)=-1, then differentiate −1-1 to get tangent gradient zero. Evaluating first loses the changing function. Right: differentiate f(x)f(x), then calculate f′(2)=1f'(2)=1.

Change the input once more and obtain the new tangent.

Normals and horizontal tangents

The normal passes through the same point as the tangent and is perpendicular to it. A shallow upward tangent therefore has a steep downward normal.

Normal gradient

If the tangent has finite nonzero gradient mm, its normal has gradient

mn=−1mbecausemmn=−1m_n=-\frac1m \qquad\text{because}\qquad mm_n=-1

Use y−b=mn(x−a)y-b=m_n(x-a) through the contact point (a,b)(a,b).
Not in the formula booklet: learn it.

For the root curve below, the tangent gradient at (4,2)(4,2) is 1/41/4, so the normal gradient is −4-4.

The square-root curve with its tangent and perpendicular normal at four two

The negative reciprocal changes both sign and magnitude. It applies only when both gradients are finite and the tangent gradient is nonzero.

Worked example 3: normal to a root curve

Question. Find the normal to y=xy=\sqrt x at x=4x=4.

Plan. Find the contact point, evaluate the derivative and take its negative reciprocal.

f(4)=2heightf′(x)=12xf′(4)=14tangent gradientmn=−4negative reciprocaly−2=−4(x−4)normal through the point\begin{aligned} f(4)&=2 && \text{height}\\ f'(x)&=\frac1{2\sqrt x}\qquad f'(4)=\frac14 && \text{tangent gradient}\\ m_n&=-4 && \text{negative reciprocal}\\ y-2&=-4(x-4) && \text{normal through the point} \end{aligned}

Why this step? The normal uses the tangent's gradient to find its direction, but still uses the original curve for its point.

Answer. y−2=−4(x−4)y-2=-4(x-4).

Check. The line passes through (4,2)(4,2) and (1/4)(−4)=−1(1/4)(-4)=-1, confirming perpendicular gradients.

Change only the contact input to 99. The tangent becomes shallower, so the normal becomes steeper.

Worked example 4: a new contact point

Question. Find the normal to y=xy=\sqrt x at x=9x=9.

Plan. Use the same derivative, evaluated at the new point, and construct the perpendicular line.

f(9)=3heightf′(9)=12(3)=16tangent gradientmn=−6negative reciprocaly−3=−6(x−9)normal equation\begin{aligned} f(9)&=3 && \text{height}\\ f'(9)&=\frac1{2(3)}=\frac16 && \text{tangent gradient}\\ m_n&=-6 && \text{negative reciprocal}\\ y-3&=-6(x-9) && \text{normal equation} \end{aligned}

Why this step? Taking only the negative, −1/6-1/6, would change the direction without making the line perpendicular.

Answer. y−3=−6(x−9)y-3=-6(x-9).

Check. It contains (9,3)(9,3), and (1/6)(−6)=−1(1/6)(-6)=-1.

If f′(a)=0f'(a)=0, the tangent is horizontal, y=f(a)y=f(a), and the normal is vertical, x=ax=a, with undefined gradient. For y=x2−4x+7y=x^2-4x+7 at 22, these are y=3y=3 and x=2x=2.

An undefined derivative needs care. For y=x3y=\sqrt[3]x at the origin, chord gradients are h−2/3→+∞h^{-2/3}\to+\infty: there is a vertical tangent x=0x=0, whose normal is horizontal, y=0y=0. There is no finite derivative there. For y=∣x∣y=|x|, however, different finite slopes approach from the two sides, so there is a corner and no unique tangent or normal. Undefined derivative does not automatically mean vertical tangent. These special-case line rules are not supplied in the formula booklet.

Watch out

Wrong: “Tangent gradient zero gives normal gradient −1/0=0-1/0=0.” Division by zero is undefined. Right: use directions: a horizontal tangent has a vertical normal x=ax=a.

Return to the root curve and change only its contact input once more.

Increasing and decreasing intervals

An interval statement concerns every pair of ordered inputs in that interval, not just a few sample heights. A function is strictly increasing if a<ba<b implies f(a)<f(b)f(a)<f(b), and strictly decreasing if a<ba<b implies f(a)>f(b)f(a)>f(b).

Showing f′>0f'>0 throughout an open interval establishes strict increase there; f′<0f'<0 establishes strict decrease. AQA's preferred interval convention uses these strict derivative inequalities, while also accepting the convention permitting derivative zero. The pairwise definition explains why an isolated zero need not interrupt strict increase. These criteria are not supplied in the formula booklet.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account