Solved Examples: Applications of Derivatives

The gradient(slope) of the tangent line is the derivative

$ y = 2kx^2 + x + 1 \:\:at\:\:x = 1 \\[3ex] \dfrac{dy}{dx} = 4kx + 1 \\[5ex] At\:\: x = 1, \:\:\dfrac{dy}{dx} = 9 \\[5ex] \dfrac{dy}{dx}\Big|_{x = 1} = 4 * k * 1 + 1 = 9 \\[5ex] 4k + 1 = 9 \\[3ex] 4k = 9 - 1 \\[3ex] 4k = 8 \\[3ex] k = \dfrac{8}{4} \\[5ex] k = 2 $

The slope is the derivative
Power Rule

$ y = 2x^2 + 5x - 3 \\[3ex] y' = 4x + 5 \\[3ex] (1, 4) \rightarrow x = 1, y = 4 \\[3ex] y' = 4(1) + 5 \\[3ex] y' = 4 + 5 \\[3ex] y' = 9 $

$ y = 3x^2 - x + 1 \\[3ex] y' = 6x - 1 \\[3ex] \text{gradient} = y'\;\;@\;\;x = -1 \\[3ex] = 6(-1) - 1 \\[3ex] = -6 - 1 \\[3ex] = -7 $

1st: We shall find the slope of the tangent to the curve at at x = 2.
This is the derivative of the function at x = 2.

2nd: We shall determine the slope of the normal to the curve
The normal to the curve is perpendicular to the tangent to the curve
Hence, the product of the two slopes (product of the slope of the tangent and the slope of the normal) is −1

3rd: We shall determine the point (x₁, y₁) on the curve
We already know the x-coordinate, but we shall find the y-coordinate

4th: We shall use the Point-Slope Form to determine the equation of the normal to the curve.

$ y = 7x - 5x^2 \\[3ex] \underline{\text{Slope of the tangent to the curve}}, m_T \\[3ex] \dfrac{dy}{dx} = 7 - 10x \\[3ex] @\;x = 2 \\[3ex] m_T = \left.\dfrac{dy}{dx}\right|_{x = 2} \\[5ex] = 7 - 10(2) \\[3ex] = 7 - 20 \\[3ex] = -13 \\[5ex] \underline{\text{Slope of the normal to the curve}}, m_N \\[3ex] m_N * m_T = -1 \\[3ex] m_N * 13 = -1 \\[3ex] m_N = \dfrac{-1}{-13} \\[5ex] m_N = \dfrac{1}{13} \\[5ex] \underline{\text{Point on the curve}}, (x_1, y_1) \\[3ex] y = 7x - 5x^2 \\[3ex] @\;x = 2 \\[3ex] y = 7(2) - 5(2)^2 \\[3ex] = 14 - 5(4) \\[3ex] = 14 - 20 \\[3ex] = -6 \\[3ex] (x_1, y_1) = (2, -6) \\[5ex] \underline{\text{Equation of the normal to the curve}} \\[3ex] \text{Point – Slope Form} \\[3ex] y - y_1 = m_N(x - x_1) \\[4ex] y - -6 = \dfrac{1}{13}(x - 2) \\[5ex] y + 6 = \dfrac{1}{13}x - \dfrac{2}{13} \\[5ex] y = \dfrac{1}{13}x - \dfrac{2}{13} - 6 \\[5ex] y = \dfrac{1}{13}x - \dfrac{2}{13} - \dfrac{78}{13} \\[5ex] y = \dfrac{1}{13}x - \dfrac{80}{13} \\[5ex] LCD = 13 \\[3ex] \text{Multiply each term by 13} \\[3ex] 13y = 13\left(\dfrac{1}{13}x\right) - 13\left(\dfrac{80}{13}\right) \\[5ex] 13y = x - 80 \\[3ex] 13y - x + 80 = 0 $

$ f(\theta) = \dfrac{4}{6 + 2\cos\theta} \\[5ex] u = 4 \\[3ex] \dfrac{du}{d\theta} = 0 \\[5ex] v = 6 + 2\cos\theta \\[3ex] \dfrac{dv}{d\theta} = 2 * -\sin\theta + \cos\theta * 0 = -2\sin\theta + 0 = -2\sin\theta \\[5ex] f'(\theta) = \dfrac{v\dfrac{du}{d\theta} - u\dfrac{du}{d\theta}}{v^2} \\[5ex] f'(\theta) = \dfrac{(6 + 2\cos\theta)(0) - 4(-2\sin\theta)}{(6 + 2\cos\theta)^2} \\[5ex] f'(\theta) = \dfrac{-4(-2\sin\theta)}{(6 + 2\cos\theta)^2} \\[5ex] Set\:\:f'(\theta) = 0 \\[3ex] \dfrac{-4(-2\sin\theta)}{(6 + 2\cos\theta)^2} = 0 \\[5ex] -4(-2\sin\theta) = 0((6 + 2\cos\theta)^2) \\[3ex] 8\sin\theta = 0 \\[3ex] \sin\theta = \dfrac{0}{8} \\[5ex] \sin\theta = 0 \\[3ex] \theta = 0, \pi, 2\pi ...Unit\:\:Circle\:\:Trigonometry \\[3ex] But\:\:\dfrac{\pi}{2} \lt \theta \lt 2\pi...Question \\[3ex] \therefore \theta = \pi = 180^\circ \\[3ex] Substitute\:\:for\:\:\theta\:\:in\:\:the\:\:main\:\:equation \\[3ex] f(\theta) = \dfrac{4}{6 + 2\cos180^\circ} \\[5ex] \cos180^\circ = -1 \\[3ex] f(\theta) = \dfrac{4}{6 + 2(-1)} \\[5ex] f(\theta) = \dfrac{4}{6 - 2} \\[5ex] f(\theta) = \dfrac{4}{4} \\[5ex] f(\theta) = 1 $

$ y = 3x^2 + 2x - 7 \\[3ex] f(x) = 3x^2 + 2x - 7 \\[3ex] (a.) \\[3ex] \dfrac{dy}{dx} = 6x + 2 \\[5ex] (b.) \\[3ex] \underline{\text{Second Derivative Test}} \\[3ex] \text{Critical Points: Set the derivative to zero and solve for }x \\[3ex] 6x + 2 = 0 \\[3ex] 6x = -2 \\[3ex] x = -\dfrac{2}{6} = -\dfrac{1}{3} \\[5ex] \text{Find the function value at the critical point} \\[3ex] f\left(-\dfrac{1}{3}\right) \\[5ex] = 3\left(-\dfrac{1}{3}\right)^2 + 2\left(-\dfrac{1}{3}\right) - 7 \\[5ex] = 3\left(-\dfrac{1}{9}\right) - \dfrac{2}{3} - \dfrac{21}{3} \\[5ex] = \dfrac{1}{3} - \dfrac{2}{3} - \dfrac{21}{3} \\[5ex] = \dfrac{1 - 2 - 21}{3} \\[5ex] = -\dfrac{22}{3} \\[5ex] \text{Point } \left(-\dfrac{1}{3}, -\dfrac{22}{3}\right) \text{ is a local maximum or a local minimum} \\[5ex] \text{To determine if the point is a local maximum or a local minimum:} \\[3ex] \text{Second Derivative: Find the derivative of the derivative} \\[3ex] \dfrac{d^2y}{dx^2} = 6 \\[5ex] 6 \gt 0 \\[3ex] \text{Hence, Point } \left(-\dfrac{1}{3}, -\dfrac{22}{3}\right) \text{ is a local minimum} \\[5ex] \text{To confirm if the point is a local maximum or a local minimum:} \\[3ex] \text{Critical Value, } x = -\dfrac{1}{3} \\[5ex] \text{Test Intervals are: } x \lt -\dfrac{1}{3} \hspace{1em}\text{and}\hspace{1em} x \gt -\dfrac{1}{3} \\[5ex] \text{First Derivative Test: Sign Test} \\[3ex] $

$\dfrac{dy}{dx} = 6x + 2$

$x \lt -\dfrac{1}{3}$ Let x = −1	$x \gt -\dfrac{1}{3}$ Let x = 0
$ 6(-1) + 2 \\[3ex] -4 \\[3ex] $	$ 6(0) + 2 \\[3ex] 2 \\[3ex] $
negative	positive
Because the slope changes from negative to positive, the function changes from decreasing to increasing. This implies that $x = -\dfrac{1}{3}$ is a local minimum.

$ (a.) \\[3ex] y = 8x + \dfrac{27}{2x^2} \\[5ex] = 8x + \dfrac{27}{2} * \dfrac{1}{x^2} \\[5ex] = 8x^1 + \dfrac{27}{2} * x^{-2} \\[5ex] \dfrac{dy}{dx} = 1 * 8x^{1 - 1} + -2 * \dfrac{27}{2} * x^{-2 - 1} \\[5ex] = 8x^0 + -27x^{-3} \\[3ex] = 8(1) - \dfrac{27}{x^3} \\[5ex] = 8 - \dfrac{27}{x^3} \\[5ex] (b.) \\[3ex] \dfrac{dy}{dx} = 0 \\[5ex] 8 - \dfrac{27}{x^3} = 0 \\[5ex] 8 = \dfrac{27}{x^3} \\[5ex] 8x^3 = 27 \\[3ex] x^3 = \dfrac{27}{8} \\[5ex] x = \sqrt[3]{\dfrac{27}{8}} \\[5ex] x = \dfrac{3}{2} \\[5ex] y = 8x + \dfrac{27}{2x^2} \\[5ex] y = 8x + 27 \div 2x^2 \\[5ex] @\; x = \dfrac{3}{2} \\[5ex] y = 8\left(\dfrac{3}{2}\right) + 27 \div 2\left(\dfrac{3}{2}\right)^2 \\[5ex] = 4(3) + 27 \div \dfrac{3^2}{2} \\[5ex] = 12 + 27 \div \dfrac{9}{2} \\[5ex] = 12 + 27 * \dfrac{2}{9} \\[5ex] = 12 + 6 \\[3ex] = 18 \\[3ex] \text{Stationary Point} = (x, y) = \left(\dfrac{3}{2}, 18\right) \\[5ex] $ To determine the nature of the stationary point, let us use the Second Derivative Test
If the second derivative is positive at the stationary point, the graph of the function is concave upward, and the point is a local minimum.
If the second derivative is negative, the graph of the function is concave downward, and the point is a local maximum.

$ \underline{\text{Nature of the Stationary Point: 2nd Derivative Test}} \\[3ex] \dfrac{dy}{dx} = 8 - \dfrac{27}{x^3} \\[5ex] \dfrac{dy}{dx} = 8 - 27x^{-3} \\[5ex] \dfrac{d^2y}{dx^2} = -3 * -27 * x^{-3 - 1} \\[5ex] = 81 * x^{-4} \\[3ex] = \dfrac{81}{x^4} \\[5ex] = 81 \div x^4 \\[3ex] @\; x = \dfrac{3}{2} \\[5ex] \dfrac{d^2y}{dx^2} = 81 \div \left(\dfrac{3}{2}\right)^4 \\[5ex] = 81 \div \dfrac{3^4}{2^4} \\[5ex] = 81 \div \dfrac{81}{16} \\[5ex] = 81 * \dfrac{16}{81} \\[5ex] = 16 \\[3ex] 16 \gt 0 \\[3ex] \therefore \left(\dfrac{3}{2}, 18\right) \text{is a minimum point} \\[5ex] $ Let:
m₁ = slope of the tangent to the curve = derivative
m₂ = slope of the normal to the curve
The product of the two slopes is −1 because the normal to the curve is perpendicular to the tangent to the curve

$ (c.) \\[3ex] \dfrac{dy}{dx} = 8 - \dfrac{27}{x^3} \\[5ex] @\; (2, 2) \\[3ex] m_1 = \left.\dfrac{dy}{dx}\right|_{x = 2} = 8 - \dfrac{27}{2^3} \\[5ex] = 8 - \dfrac{27}{8} \\[5ex] = \dfrac{64}{8} - \dfrac{27}{8} \\[5ex] = \dfrac{64 - 27}{8} \\[5ex] = \dfrac{37}{8} \\[5ex] m_1 * m_2 = -1 \\[3ex] m_2 = -\dfrac{1}{m_1} \\[5ex] = -1 \div m_1 \\[3ex] = -1 \div \dfrac{37}{8} \\[5ex] = -1 * \dfrac{8}{37} \\[5ex] = -\dfrac{8}{27} \\[5ex] For\;\;(2, 2) \\[3ex] (x_1, y_1) = (2, 2) \\[3ex] x_1 = 2 \\[3ex] y_1 = 2 \\[5ex] \underline{\text{Point — Slope Form}} \\[3ex] y - y_1 = m_2(x - x_1) \\[3ex] y - 2 = -\dfrac{8}{27}(x - 2) \\[5ex] y = -\dfrac{8x}{37} + \dfrac{16}{37} + 2 \\[5ex] y = -\dfrac{8x}{37} + \dfrac{16}{37} + \dfrac{74}{37} \\[5ex] y = \dfrac{-8x}{37} + \dfrac{90}{37} \\[5ex] LCD = 37 \\[3ex] 37y = 37\left(\dfrac{-8x}{37}\right) + 37\left(\dfrac{90}{37}\right) \\[5ex] 37y = -8x + 90 \\[3ex] 37y + 8x - 90 = 0 $

$ y = (x^2 - 1)x \\[3ex] y = x(x^2 - 1) \\[3ex] y' = x(2x) + (x^2 - 1)(1)...\text{Product Rule} \\[3ex] y' = 2x^2 + x^2 - 1 \\[3ex] y' = 3x^2 - 1 \\[3ex] \text{At the point } (1, 0) \\[3ex] x = 1 \\[3ex] y = 0 \\[3ex] m = y' = 3(1)^2 - 1 \\[3ex] m = 3(1) - 1 \\[3ex] m = 3 - 1 \\[3ex] m = 2 \\[3ex] \text{Equation of the tangent to the curve} \\[3ex] y - y_1 = m(x - x_1)...\text{Point-Slope Form} \\[3ex] x_1 = 1 \\[3ex] y_1 = 0 \\[3ex] m = 2 \\[3ex] \implies \\[3ex] y - 0 = 2(x - 1) \\[3ex] y = 2(x - 1) \\[3ex] y = 2x - 2 \\[3ex] $ None of the above

$ y = 3x^2 - x^3 \\[3ex] y' = 6x - 3x^2 \\[3ex] Set\:\:y = 0 \\[3ex] 6x - 3x^2 = 0 \\[3ex] 3x(2 - x) = 0 \\[3ex] 3x = 0 \:\:OR\:\: 2 - x = 0 \\[3ex] x = \dfrac{0}{3} \:\:OR\:\: 2 - 0 = x \\[5ex] x = 0 \:\:OR\:\: x = 2 \\[3ex] $ To find the maximum value of $y$ in this case (we have two values of $x$), we can solve it usint at least two approaches.
Use whichever way you prefer.

$ \text{First Approach: Second Derivative Test} \\[3ex] y' = 6x - 3x^2 \\[3ex] y'' = 6 - 6x = 6(1 - x) \\[3ex] For\:\:x = 0 \\[3ex] y'' = 6(1 - 0) = 6(1) = 6 \\[3ex] 6 \gt 0 \\[3ex] x = 0 \:\:gives\:\:minimum\:\:y-value \\[3ex] For\:\:x = 2 \\[3ex] y'' = 6(1 - 2) = 6(-1) = -6 \\[3ex] -6 \lt 0 \\[3ex] x = 2 \text{ gives maximum } y-value \\[3ex] \text{Substitute }x = 2 \text{ in the main equation} \\[3ex] y = 3(2)^2 - 2^3 \\[3ex] y = 3(4) - 8 \\[3ex] y = 12 - 8 \\[3ex] y = 4 \\[5ex] \text{Second Approach: Test Values} \\[3ex] For\:\:x = 0 \\[3ex] y = 3(0)^2 - 0^3 \\[3ex] y = 3(0) - 0 \\[3ex] y = 0 - 0 \\[3ex] y = 0 \\[3ex] For\:\:x = 2 \\[3ex] y = 3(2)^2 - 2^3 \\[3ex] y = 3(4) - 8 \\[3ex] y = 12 - 8 \\[3ex] y = 4 \\[5ex] 4 \gt 0 \\[3ex] \therefore y = 4 $

$ 2\sin x = x \\[3ex] 2\sin x - x = 0 \\[3ex] 2\sin x - x = f(x) \\[3ex] f(x) = 2\sin x - x \\[3ex] f'(x) = 2\cos x - 1 \\[3ex] x_{n + 1} = x_n - \dfrac{f(x)}{f'(x)} \\[5ex] \implies \\[3ex] x_{n + 1} = x_n - \dfrac{2\sin x_n - x_n}{2\cos x_n - 1} \\[5ex] $

$n$	$1$	$2$
$x_n$	$2$	$1.900995594$
$\sin x_n$	$0.9092974268$	$0.9459777535$
$2\sin x_n$	$1.818594854$	$1.891955507$
$\color{purple}{2\sin x_n - x_n}$	$-0.1814051463$	$-0.009040087$
$\cos x_n$	$-0.4161468365$	$-0.3242315372$
$2\cos x_n$	$-0.8322936731$	$-0.6484630743$
$\color{purple}{2\cos x_n - 1}$	$-1.832293673$	$-1.648463074$
$\dfrac{2\sin x_n - x_n}{2\cos x_n - 1}$	$0.0990044058$	$0.0054839487$
$\color{darkblue}{x_n - \dfrac{2\sin x_n - x_n}{2\cos x_n - 1}}$	$\color{darkblue}{1.900995594}$	$\color{darkblue}{1.895511645}$

$ x_2 = 1.900995594 \\[3ex] x_3 = 1.895511645 $

$ y = \dfrac{2x}{x^2 + 1} = \dfrac{p}{k} \\[5ex] p = 2x \\[3ex] p' = 2 \\[3ex] k = x^2 + 1 \\[3ex] k' = 2x \\[3ex] y' = \dfrac{kp' - pk'}{k^2} \\[5ex] y' = \dfrac{(x^2 + 1)(2) - 2x(2x)}{(x^2 + 1)^2} \\[5ex] y' = \dfrac{2x^2 + 2 - 4x^2}{(x^2 + 1)^2} \\[5ex] y' = \dfrac{2 - 2x^2}{(x^2 + 1)^2} \\[5ex] \text{Stationary Points} \\[3ex] y' = 0 \\[3ex] \implies \\[3ex] \dfrac{2 - 2x^2}{(x^2 + 1)^2} = 0 \\[5ex] 2 - 2x^2 = 0 \\[3ex] 2 = 2x^2 \\[3ex] 2x^2 = 2 \\[3ex] x^2 = \dfrac{2}{2} \\[5ex] x^2 = 1 \\[3ex] x = \pm \sqrt{1} \\[3ex] x = \pm 1 \\[3ex] y = \dfrac{2x}{x^2 + 1} \\[5ex] when\;\; x = 1 \\[3ex] y = \dfrac{2(1)}{1^2 + 1} \\[5ex] y = \dfrac{2}{1 + 1} \\[5ex] y = \dfrac{2}{2} \\[5ex] y = 1 \\[3ex] (x, y) \rightarrow (1, 1) \\[3ex] when\;\; x = -1 \\[3ex] y = \dfrac{2(-1)}{(-1)^2 + 1} \\[5ex] y = \dfrac{-2}{1 + 1} \\[5ex] y = \dfrac{-2}{2} \\[5ex] y = -1 \\[3ex] (x, y) \rightarrow (-1, -1) \\[3ex] $ The stationary points are: (1, 1) and (−1, −1)

$ (25.1) \\[3ex] Function:\;\;h(x) = ax^3 + bx^2 \\[3ex] For\;\;Point\;B(4, 32) \\[3ex] x = 4 \\[3ex] y = h(x) = 32 \\[3ex] \implies \\[3ex] 32 = a(4)^3 + b(4)^2 \\[3ex] 32 = 64a + 16b \\[3ex] 64a + 16b = 32 \\[3ex] \implies \\[3ex] 4a + b = 2 ...eqn.(1) \\[3ex] Also: \\[3ex] h(x) = ax^3 + bx^2 \\[3ex] h'(x) = 3ax^2 + 2bx \\[3ex] At\;\;turning\;\;point,\;\;h'(x) = 0 \\[3ex] For\;\;Turning\;\;Point\;B(4, 32) \\[3ex] x = 4 \\[3ex] \implies \\[3ex] h'(4) = 0 \\[3ex] h'(4) = 3a(4)^2 + 2b(4) = 0 \\[3ex] 48a + 8b = 0 \\[3ex] \implies \\[3ex] 6a + b = 0 ...eqn.(2) \\[3ex] eqn.(2) - eqn.(1) \\[3ex] 6a - 4a = 0 - 2 \\[3ex] 2a = -2 \\[3ex] a = -\dfrac{2}{2} \\[5ex] a = -1 \\[3ex] Substitute\;\;a = -1 \;\;into\;\;eqn.(1) \\[3ex] 4(-1) + b = 2 \\[3ex] -4 + b = 2 \\[3ex] b = 2 + 4 \\[3ex] b = 6 \\[3ex] \therefore h(x) = -x^3 + 6x^2 \\[3ex] (25.2) \\[3ex] A = x-intercept\;\;of\;\;h(x) \\[3ex] \implies \\[3ex] h(x) = 0 \\[3ex] -x^3 + 6x^2 = 0 \\[3ex] x^2(-x + 6) = 0 \\[3ex] x^2 = 0 \;\;\;OR\;\;\; -x + 6 = 0 \\[3ex] x = \pm \sqrt{0} \;\;\;OR\;\;\; 6 = 0 + x \\[3ex] x = \pm 0 \;\;\;OR\;\;\; 6 = x \\[3ex] Based\;\;on\;\;the\;\;graph:\;\; x \ne \pm 0 \\[3ex] x = 6 \\[3ex] \therefore coordinates\;\;of\;\;A = (6, 0) \\[3ex] (25.3) \\[3ex] (25.3.1) \\[3ex] Based\;\;on\;\;the\;\;Graph \\[3ex] h(x)\uparrow for\;\; x \in (0, 4) \\[3ex] (25.3.2) \\[3ex] Based\;\;on\;\;the\;\;function \\[3ex] h(x) = -x^3 + 6x^2 \\[3ex] h'(x) = -3x^2 + 12x \\[3ex] h''(x) = -6x + 12 \\[3ex] h''(x) = 0 \\[3ex] \implies \\[3ex] -6x + 12 = 0 \\[3ex] -6x = -12 \\[3ex] x = \dfrac{-12}{-6} \\[5ex] Inflection\;\;Point:\;\;x = 2 \\[3ex] Intervals:\;\; x \lt 2 \;\;and\;\; x \gt 2 \\[3ex] $

$h''(x) = -6x + 12 $

Interval	$x \lt 2$	$x \gt 2$
Test Value	$x = 1$	$x = 3$
Sign Test	$$ -6(1) + 12 \\[3ex] -6 + 12 = 6 \\[3ex] 6 \gt 0 \\[3ex] positive $$	$$ -6(3) + 12 \\[3ex] -18 + 12 = -6 \\[3ex] -6 \lt 0 \\[3ex] negative $$
Conclusion	Concave up	Concave down

$ h(x)\frown for\;\; x \gt 2 \\[3ex] h(x)\frown for\;\; x \in (2, \infty) \\[3ex] (25.4) \\[3ex] -(x - 1)^3 + 6(x - 1)^2 - k = 0 \\[3ex] -(x - 1)^3 + 6(x - 1)^2 = k \\[3ex] k = -(x - 1)^3 + 6(x - 1)^2 \\[3ex] h(x) = -x^3 + 6x^2 \\[3ex] h(x - 1) = -(x - 1)^3 + 6(x - 1)^2 \\[3ex] \implies k = h(x - 1) \\[3ex] k = graph\;\;of\;\;h(x - 1) \\[3ex] 1st\;\;find\;\;the\;\;y-intercept...value\;\;of\;\;k\;\;when\;\;x = 0 \\[3ex] k = -(0 - 1)^3 + 6(0 - 1)^2 \\[3ex] k = -(-1)^3 + 6(-1)^2 \\[3ex] k = -(-1) + 6(1) \\[3ex] k = 1 + 6 \\[3ex] k = 7 \\[3ex] Negative\;\;root \implies x \lt 0 \\[3ex] When\;\;x \lt 0,\;\;k \lt 7 \\[3ex] Let\;\;us\;\;find\;\;the\;\;maximum\;\;value\;\;for\;\;the\;\;positive\;\;roots \\[3ex] h(x - 1) = -(x - 1)^3 + 6(x - 1)^2 \\[3ex] h(x - 1)' = -3(x - 1)^2 + 12(x - 1) \\[3ex] h(x - 1)' = 0 \\[3ex] \implies \\[3ex] -3(x - 1)^2 + 12(x - 1) = 0 \\[3ex] -3(x^2 - 2x + 1) + 12x - 12 = 0 \\[3ex] -3x^2 + 6x - 3 + 12x - 12 = 0 \\[3ex] -3x^2 + 18x - 15 = 0 \\[3ex] x^2 - 6x + 5 = 0 \\[3ex] (x - 5)(x - 1) = 0 \\[3ex] x - 5 = 0 \;\;\;OR\;\;\; x - 1 = 0 \\[3ex] x = 5 \;\;\;OR\;\;\; x = 1 \\[3ex] For\;\;x = 5 \\[3ex] h(x - 1) = -(5 - 1)^3 + 6(5 - 1)^2 \\[3ex] = -4^3 + 6(4^2) \\[3ex] = -64 + 6(16) \\[3ex] = -64 + 96 \\[3ex] = 32 \\[3ex] For\;\;x = 1 \\[3ex] h(x - 1) = -(1 - 1)^3 + 6(0 - 1)^2 \\[3ex] = -0^3 + 6(-1)^2 \\[3ex] = 0 + 6(1) \\[3ex] = 0 + 6 \\[3ex] = 6 \\[3ex] Maximum\;\;value = 32 \\[3ex] \implies \\[3ex] 7 \lt k \lt 32 $

$ (a.) \\[3ex] f(x) = \dfrac{7}{x} - 7\ln(x) \\[5ex] f(x) = 7x^{-1} - 7\ln x \\[3ex] f'(x) = -7x^-2 - \left[7\left(\dfrac{1}{x}\right) + \ln x(0)\right] \\[5ex] f'(x) = -7x^{-2} - \dfrac{7}{x} \\[5ex] f'(x) = \dfrac{-7}{x^2} - \dfrac{7}{x} \\[5ex] Simplify\;\;f(x)\;\;and\;\;f'(x) \\[3ex] f(x) = \dfrac{7}{x} - 7\ln(x) \\[5ex] f(x) = \dfrac{7 - 7x\ln x}{x} \\[5ex] f'(x) = \dfrac{-7}{x^2} - \dfrac{7}{x} \\[5ex] f'(x) = \dfrac{-7 - 7x}{x^2} \\[5ex] \dfrac{f(x)}{f'(x)} \\[5ex] = f(x) \div f'(x) \\[3ex] = \dfrac{7 - 7x\ln x}{x} \div \dfrac{-7 - 7x}{x^2} \\[5ex] = \dfrac{7 - 7x\ln x}{x} * \dfrac{x^2}{-7 - 7x} \\[5ex] = \dfrac{7(1 - x\ln x)}{x} * \dfrac{x^2}{-7(1 + x)} \\[5ex] = \dfrac{x(1 - x\ln x)}{-(1 + x)} \\[5ex] = \dfrac{x - x^2\ln x}{-(1 + x)} \\[5ex] = -\dfrac{(x - x^2\ln x)}{1 + x} \\[5ex] = \dfrac{-x + x^2\ln x}{x + 1} \\[5ex] = \dfrac{x^2\ln x - x}{x + 1} \\[5ex] x - \dfrac{f(x)}{f'(x)} \\[5ex] = x - \dfrac{x^2\ln x - x}{x + 1} \\[5ex] = \dfrac{x(x + 1) - (x^2\ln x - x)}{x + 1} \\[5ex] = \dfrac{x^2 + x - x^2\ln x + x}{x + 1} \\[5ex] = \dfrac{x^2 + 2x - x^2\ln x}{x + 1} \\[5ex] x_{n + 1} = x_n - \dfrac{f(x)}{f'(x)} \\[5ex] \implies \\[3ex] x_{n + 1} = \dfrac{x_n^2 + 2x_n - x_n^2\ln x_n}{x_n + 1} \\[5ex] x_1 = 1 \\[3ex] \implies \\[3ex] x_2 = \dfrac{x_1^2 + 2x_1 - x_1^2\ln x_1}{x_1 + 1} \\[5ex] x_2 = \dfrac{1^2 + 2(1) - 1^2 * \ln(1)}{1 + 1} \\[5ex] x_2 = \dfrac{1 + 2 - 1 * 0}{2} \\[5ex] x_2 = \dfrac{1 + 2 - 0}{2} \\[5ex] x_2 = \dfrac{3}{2} \\[7ex] x_3 = \dfrac{x_2^2 + 2x_2 - x_2^2\ln x_2}{x_2 + 1} \\[5ex] \underline{Numerator} \\[3ex] x_2^2 + 2x_2 - x_2^2\ln x_2 \\[3ex] = \left(\dfrac{3}{2}\right)^2 + 2 * \dfrac{3}{2} - \left(\dfrac{3}{2}\right)^2 * \ln \left(\dfrac{3}{2}\right) \\[5ex] = \dfrac{9}{4} + 3 - \dfrac{9}{4} * \ln \left(\dfrac{3}{2}\right) \\[5ex] = \dfrac{21}{4} - \dfrac{9}{4} * \ln \left(\dfrac{3}{2}\right) \\[5ex] \underline{Denominator} \\[3ex] x_2 + 1 \\[3ex] = \dfrac{3}{2} + 1 \\[5ex] = \dfrac{3 + 2}{2} \\[5ex] = \dfrac{5}{2} \\[5ex] Numerator \div Denominator \\[3ex] = \dfrac{21}{4} - \dfrac{9}{4} * \ln \left(\dfrac{3}{2}\right) \div \dfrac{5}{2} \\[5ex] = \dfrac{1}{4}\left[21 - 9\ln\left(\dfrac{3}{2}\right)\right] * \dfrac{2}{5} \\[5ex] = \dfrac{1}{10}\left[21 - 9\ln\left(\dfrac{3}{2}\right)\right] \\[5ex] = \dfrac{21}{10} - \dfrac{9}{10}\ln\left(\dfrac{3}{2}\right) \\[5ex] \therefore x_3 = \dfrac{21}{10} - \dfrac{9}{10}\ln\left(\dfrac{3}{2}\right) \\[5ex] (b.) \\[3ex] x_3 = 2.1 - (0.9 * \ln(1.5)) \\[3ex] x_3 = 1.735081403 \\[3ex] $



$ x_4 = 1.762915391 \\[3ex] $
$ x_5 = 1.763222798 \\[3ex] x_6 = 1.763222834 \\[3ex] Because\;\;x_5 \approx x_6...STOP $