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

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

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

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

• \(\frac{dy}{dx}=3 x^{2} + 2 a x\)
• At x = −3: \(-6 a + 27=0\)
• a = 4.5

The curve \(y=a x^{2} + \frac{12}{x}\) has a horizontal tangent at x = 2. Find the value of a.

• \(\frac{dy}{dx}=2 a x - \frac{12}{x^{2}}\)
• At x = 2: \(-3 + 4 a=0\)
• a = 0.75

The curve \(y=\frac{a}{x^{2}}-6x\) has a horizontal tangent at x = −2. Find the value of a.

• \(\frac{dy}{dx}=-6 - \frac{2 a}{x^{3}}\)
• At x = −2: \(-6 + \frac{a}{4}=0\)
• a = 24

Solve \(f'(x)=0\) for \(f(x)=x^{3} - 3 x^{2} - 24 x + 5\).

• Graph \(y=\frac{d}{dx}\left(x^{3} - 3 x^{2} - 24 x + 5\right)\)
• Find where the graph meets the x-axis (the zeros)
• x = −2 or 4

Solve \(f'(x)=0\) for \(f(x)=x^{4} - 6 x^{2} + 4 x\).

• Graph \(y=\frac{d}{dx}\left(x^{4} - 6 x^{2} + 4 x\right)\)
• Find where the graph meets the x-axis (the zeros)
• x = −1.88 or 0.347 or 1.53

Solve \(f'(x)=0\) for \(f(x)=5 x + \frac{20}{x^{2}}\).

• Graph \(y=\frac{d}{dx}\left(5 x + \frac{20}{x^{2}}\right)\)
• Find where the graph meets the x-axis (the zeros)
• x = 2

A tour operator models its daily profit, \(P\) hundred euros, when it charges x euros for a ticket as \(P=40x-ax^2\). The profit is greatest when the price is 8 euros. Find the value of a.

• \(\frac{dP}{dx}=-2 a x + 40\)
• Greatest at x = 8, so the gradient is 0: \(-16 a + 40=0\)
• a = 2.5

The concentration of a drug in the blood, \(c\) mg per litre, is modelled by \(c=t^3-15t^2+63t\), where t is the time in hours after an injection, \(0\le t\le 10\). Use your calculator to find the times when the concentration is neither rising nor falling.

• Graph \(y=\frac{d}{dt}\left(t^{3} - 15 t^{2} + 63 t\right)\) for \(0\le t\le 10\)
• From the calculator: the zeros of the graph
• t = 3 and t = 7 hours

The curve \(y=x^3+ax^2+bx\) has horizontal tangents at x = 1 and at x = 5. 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 = 5: \(10 a + 75 + b=0\)
• Solve the two equations together (simultaneous equations on the calculator)
• a = −9, b = 15

The height of a path above a valley floor, \(h\) metres, at a horizontal distance of x hundred metres is \(h=\frac{1}{2}x^4-4x^3+6x^2\). Find the values of x, to 3 significant figures, where the path is level (the gradient is zero).

• Graph \(y=\frac{d}{dx}\left(\frac{x^{4}}{2} - 4 x^{3} + 6 x^{2}\right)\)
• From the calculator: the zeros of the graph
• x = 0, 1.27, 4.73