###### 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 $$t = 2$$$$\frac{\mathrm dy}{\mathrm dx} = -\frac{3\times2^5 + 2\times 2^3}{2} = -3\times 16 - 8 = -56$$.