4.1

The curve \(y = (1 - x)(x^2 + 4x + k)\) has a stationary point when \(x = -3\).

Find the value of the constant \(k\).

\(y = (1 - x)(x^2 + 4x + k) = x^2 + 4x + k - x^3 - 4x^2 - kx = -x^3 - 3x^2 + (4 - k)x + k\). \(y' = -3x^2 - 6x + 4 - k\). Since \(x = -3\) is a stationary point, \(y'(-3) = 0\), so \(-3 - 6 + 4 - k = 0\). Therefore, \(k = -5\).

4.2

Find \(\frac{dy}{dx}\) of \(y = \frac{(3x)^2 \times x^4}{x}\). Express the answer in the form \(ax^5\).

Find \(a\).

\(\frac{(3x)^2 \times x^4}{x} = \frac{9x^2 \times x^4}{x} = 9x^6\) 

\(\frac{d}{dx}(9x^6) = 9\times 6x^5 = 54x^5\)

4.3

Given the curve \(y = x^4 + 32x\).

Find the coordinates of the stationary point on the curve.

\(y' = 4x^3 + 32\). To find the stationary point we should solve the equation \(y' = 0\):

\(4x^3 + 32 = 0,\\ 4x^3 = -32,\\ x^3 = -8,\\ x = -2.\)

\(y(-2) = (-2)^4 + 32 \times (-2) = 16 - 64 = -48\).

4.4

Given that \(y = (x + 2)(x^2 - 3x + 5)\).

Find the minimum value of the function \(y(x)\).

\(y = (x + 2)(x^2 - 3x + 5) = x^3 - 3x^2 + 5x + 2x^2 - 6x + 10 = x^3 - x^2 - x + 10\). \(y' = 3x^2 - 2x - 1\). Solve the equation \(y' = 0\): \(3x^2 - 2x - 1 = 0\). Discriminant \(D = b^2 - 4ac = (-2)^2 - 4\times3\times(-1) = 4 + 12 = 16\). \(x = \frac{-b \pm \sqrt{D}}{2a} = \frac{2 \pm \sqrt{16}}{2\times3} = \frac{2 \pm 4}{6}\). So, \(x = 1\) or \(-\frac{1}{3}\). Test \(x = 1\) as a candidate for the minimum point. Choose two points from the left and the right side of the \(x = 1\), for example, \(0\) and \(2\). \(y'(0) = -1 < 0\), \(y' (2) = 12 - 4 - 1 = 7 > 0\). Therefore, \(y(x)\) decreases before \(x = 1\) and increases after \(x = 1\), so \(x = 1\) is the minimum point. The minimum value of the \(y(x)\) is \(y(1) = (1 + 2)(1 - 3 + 5) = 3\times3 = 9\).

4.5

Given that \(f(x) = \frac{4}{x} - 3x + 2\).

 Find \(f''(\frac{1}{2})\).

\(f'(x) = (4x^{-1})' - 3 = -4x^{-2} - 3\). \(f''(x) = -4\times (-2)x^{-3} = \frac{8}{x^3}\). \(f''(\frac{1}{2}) = \frac{8}{\frac{1}{8}} = 8 \times 8 = 64\).

4.6

Given \(f(x) = (25x^{-2})^{-\frac{1}{2}}\).

Find \(f'(x)\).

\(f(x) = (25x^{-2})^{-\frac{1}{2}} = \frac{1}{\sqrt{25x^{-2}}} = \frac{1}{5x^{-1}} = \frac{x}{5}\), so, \(f'(x) = \frac{1}{5} = 0.2\).

4.7

Given the curve \(y = 3x^2 - \frac{6}{x} - 2\).

Find the coordinates of the stationary point on the curve.

\(y' = 6x + \frac{6}{x^2}\). Solve the equation \(y' = 0\):

\(6x + \frac{6}{x^2} = 0,\\ 6x = -\frac{6}{x^2},\\ x^3 = -1,\\ x = -1.\)

So, \(x = -1\) is a stationary point. \(y(-1) = 6\times (-1)^2 - \frac{6}{-1} - 2 = 6 + 6 - 2 = 10\).

4.8

Given that \(f(x) = \frac{1}{2x^2} - \frac{5}{x} + 2x\).

 Find \(f'(-1)\).

\(f(x) = \frac{1}{2}x^{-2} - 5x^{-1} + 2x\), \(f'(x) = \frac{1}{2}\times(-2)x^{-3} - 5\times(-1)x^{-2} + 2 = -\frac{1}{x^3} + \frac{5}{x^2} + 2\), \(f'(-1) = -\frac{1}{(-1)^3} + \frac{5}{(-1)^2} + 2 = 1 + 5 + 2 = 8\).

4.9

Given the curve \(y = 3x^2 + 18x - 7\).

Find the coordinates of the stationary point on the curve.

\(y' = 6x + 18\). Stationary point is the root of the equation \(y' = 0\):

\(6x + 18 = 0,\\ 6x = -18,\\ x = -3.\).

\(y(-3) = 3\times(-3)^2 + 18\times(-3) - 7 = 27 - 54 - 7 = -34\).

4.10

Find \(\frac{d^2 y}{dx^2}\) of \(y = 5x - 2 + \frac{5}{x^2}\). Express the answer in the form \(\frac{k}{x^4}\).

Find \(k\).

\(\frac{dy}{dx} = 5 + \frac{d}{dx}(5x^{-2}) = 5 - 10x^{-3}\), \(\frac{d^2 y}{dx^2} = \frac{d}{dx}(5 - 10x^{-3}) = 30x^{-4} = \frac{30}{x^4}\).

4.11

A curve is defined by the parametric equations \(x = \frac{1}{t^2}+5\)\(y = t^3 +2t-1\).

Find the gradient at the point on the curve where \(t = 2\).

Since \(\frac{\mathrm dx}{\mathrm dt} = -\frac{2}{t^3}\) and \(\frac{\mathrm dy}{\mathrm dt} = 3t^2\), the gradient of the curve is equal to \(\frac{\mathrm dy}{\mathrm dx} = \frac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}} = \frac{3t^2+2}{-\frac{2}{t^3}} = -\frac{3t^5 + 2t^3}{2}\). At \(t = 2\)\(\frac{\mathrm dy}{\mathrm dx} = -\frac{3\times2^5 + 2\times 2^3}{2} = -3\times 16 - 8 = -56\).