Stationary points: where the gradient is zero Problem set B One question at a time

The curve \(y=3 x^{2} - a x\) has a horizontal tangent at x = 4. Find the value of a.

• \(\frac{dy}{dx}=6 x - a\)
• At x = 4: \(-a + 24=0\)
• a = 24

The curve \(y=x^{3} + a x^{2}\) has a horizontal tangent at x = −2. Find the value of a.

• \(\frac{dy}{dx}=3 x^{2} + 2 a x\)
• At x = −2: \(-4 a + 12=0\)
• a = 3

The curve \(y=a x^{2} + \frac{18}{x}\) has a horizontal tangent at x = 3. Find the value of a.

• \(\frac{dy}{dx}=2 a x - \frac{18}{x^{2}}\)
• At x = 3: \(-2 + 6 a=0\)
• a = 0.333

The curve \(y=\frac{a}{x^{2}}-4x\) has a horizontal tangent at x = −1. Find the value of a.

• \(\frac{dy}{dx}=-4 - \frac{2 a}{x^{3}}\)
• At x = −1: \(-4 + 2 a=0\)
• a = 2

Solve \(f'(x)=0\) for \(f(x)=x^{3} + 3 x^{2} - 9 x + 2\).

• Graph \(y=\frac{d}{dx}\left(x^{3} + 3 x^{2} - 9 x + 2\right)\)
• Find where the graph meets the x-axis (the zeros)
• x = −3 or 1

Solve \(f'(x)=0\) for \(f(x)=2 x^{4} - 5 x^{2} + x\).

• Graph \(y=\frac{d}{dx}\left(2 x^{4} - 5 x^{2} + x\right)\)
• Find where the graph meets the x-axis (the zeros)
• x = −1.17 or 0.101 or 1.06

Solve \(f'(x)=0\) for \(f(x)=4 x + \frac{54}{x^{2}}\).

• Graph \(y=\frac{d}{dx}\left(4 x + \frac{54}{x^{2}}\right)\)
• Find where the graph meets the x-axis (the zeros)
• x = 3

A farm models the yield of wheat, \(Y\) tens of kilograms per hectare, when n tens of kilograms of fertiliser are used as \(Y=60n-an^2\). The yield is greatest when n = 5. Find the value of a.

• \(\frac{dY}{dn}=-2 a n + 60\)
• Greatest at n = 5, so the gradient is 0: \(-10 a + 60=0\)
• a = 6

The depth of water in a harbour, \(d\) centimetres above a marker, is modelled by \(d=t^3-18t^2+81t\), where t is the time in hours after midnight, \(0\le t\le 12\). Use your calculator to find the times when the depth is neither rising nor falling.

• Graph \(y=\frac{d}{dt}\left(t^{3} - 18 t^{2} + 81 t\right)\) for \(0\le t\le 12\)
• From the calculator: the zeros of the graph
• t = 3 and t = 9 hours

The curve \(y=x^3+ax^2+bx\) has horizontal tangents at x = 1 and at x = 4. Find the values of a and b.

• \(\frac{dy}{dx}=3 x^{2} + 2 a x + b\)
• At x = 1: \(2 a + 3 + b=0\). At x = 4: \(48 + 8 a + b=0\)
• Solve the two equations together (simultaneous equations on the calculator)
• a = −7.5, b = 12

The height of a cycle track above ground level, \(h\) metres, at a horizontal distance of x hundred metres is \(h=x^4-6x^3+8x^2\). Find the values of x, to 3 significant figures, where the track is level (the gradient is zero).

• Graph \(y=\frac{d}{dx}\left(x^{4} - 6 x^{3} + 8 x^{2}\right)\)
• From the calculator: the zeros of the graph
• x = 0, 1.22, 3.28