Think! Additional Mathematics Textbook (10th edition) Solutions
Ex 11F
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Solutions
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(a)
\begin{align} f(x) & = x^2 + 10x + 5 \\ \\ f'(x) & = 2x + 10 \\ \\ f'(x) & > 0 \phantom{0000000000} [\text{Increasing function}] \\ 2x + 10 & > 0 \\ 2x & > - 10 \\ x & > {-10 \over 2} \\ x & > - 5 \end{align}
(b)
\begin{align} f(x) & = 3x^2 - 12x + 7 \\ \\ f'(x) & = 3(2)x - 12 \\ & = 6x - 12 \\ \\ f'(x) & > 0 \phantom{0000000000} [\text{Increasing function}] \\ 6x - 12 & > 0 \\ 6x & > 12 \\ x & > {12 \over 6} \\ x & > 2 \end{align}
(c)
\begin{align} f(x) & = 8 - 6x - 3x^2 \\ \\ f'(x) & = -6 - 3(2)x \\ & = -6 - 6x \\ \\ f'(x) & > 0 \phantom{0000000000} [\text{Increasing function}] \\ -6 - 6x & > 0 \\ -6x & > 6 \\ x & < {6 \over -6} \\ x & < - 1 \end{align}
(d)
\begin{align} f(x) & = x^3 - 12x^2 + 45x + 6 \\ \\ f'(x) & = 3x^2 - 12(2)x + 45 \\ & = 3x^2 - 24x + 45 \\ \\ f'(x) & > 0 \phantom{0000000000} [\text{Increasing function}] \\ 3x^2 - 24x + 45 & > 0 \\ x^2 - 8x + 15 & > 0 \\ (x - 3)(x - 5) & > 0 \end{align}
$$ x < 3 \text{ or } x > 5$$
(e)
\begin{align} f(x) & = 2x^3 - 3x^2 - 72x + 5 \\ \\ f'(x) & = 2(3)x^2 - 3(2)x - 72 \\ & = 6x^2 - 6x - 72 \\ \\ f'(x) & > 0 \phantom{0000000000} [\text{Increasing function}] \\ 6x^2 - 6x - 72 & > 0 \\ x^2 - x - 12 & > 0 \\ (x + 3)(x - 4) & > 0 \end{align}
$$ x < -3 \text{ or } x > 4$$
(f)
\begin{align} f(x) & = -4x^3 + 9x^2 + 30x - 7 \\ \\ f'(x) & = -4(3)x^2 + 9(2)x + 30 \\ & = -12x^2 + 18x + 30 \\ \\ f'(x) & > 0 \phantom{0000000000} [\text{Increasing function}] \\ -12x^2 + 18x + 30 & > 0 \\ 12x^2 - 18x - 30 & < 0 \\ 2x^2 - 3x - 5 & < 0 \\ (2x - 5)(x + 1) & < 0 \end{align}
$$ -1 < x < {5 \over 2} $$
(a)
\begin{align} f(x) & = 5x^2 - 10x + 3 \\ \\ f'(x) & = 5(2)x - 10 \\ & = 10x - 10 \\ \\ f'(x) & < 0 \phantom{0000000000} [\text{Decreasing function}] \\ 10x - 10 & < 0 \\ 10x & < 10 \\ x & < {10 \over 10} \\ x & < 1 \end{align}
(b)
\begin{align} f(x) & = 5 - 7x - 2x^2 \\ \\ f'(x) & = -7 - 2(2)x \\ & = - 7 - 4x \\ \\ f'(x) & < 0 \phantom{0000000000} [\text{Decreasing function}] \\ -7 - 4x & < 0 \\ -4x & < 7 \\ x & > -{7 \over 4} \end{align}
(c)
\begin{align} f(x) & = x^3 + 3x^2 - 24x + 11 \\ \\ f'(x) & = 3x^2 + 3(2)x - 24 \\ & = 3x^2 + 6x - 24 \\ \\ f'(x) & < 0 \phantom{0000000000} [\text{Decreasing function}] \\ 3x^2 + 6x - 24 & < 0 \\ x^2 + 2x - 8 & < 0 \\ (x + 4)(x - 2) & < 0 \end{align}
$$ -4 < x < 2 $$
(d)
\begin{align} f(x) & = 2x^3 - 3x^2 - 12x + 5 \\ \\ f'(x) & = 2(3)x^2 - 3(2)x - 12 \\ & = 6x^2 - 6x - 12 \\ \\ f'(x) & < 0 \phantom{0000000000} [\text{Decreasing function}] \\ 6x^2 - 6x - 12 & < 0 \\ x^2 - x - 2 & < 0 \\ (x + 1)(x - 2) & < 0 \end{align}
$$ -1 < x < 2 $$
(e)
\begin{align} f(x) & = 4x^3 - 15x^2 - 72x + 5 \\ \\ f'(x) & = 4(3)x^2 - 15(2)x - 72 \\ & = 12x^2 - 30x - 72 \\ \\ f'(x) & < 0 \phantom{0000000000} [\text{Decreasing function}] \\ 12x^2 - 30x - 72 & < 0 \\ 2x^2 - 5x - 12 & < 0 \\ (2x + 3)(x - 4) & < 0 \end{align}
$$ -{3 \over 2} < x < 4 $$
(f)
\begin{align} f(x) & = -1 + 18x + 3x^2 - 4x^3 \\ \\ f'(x) & = 18 + 3(2)x - 4(3)x^2 \\ & = 18 + 6x - 12x^2 \\ \\ f'(x) & < 0 \phantom{0000000000} [\text{Decreasing function}] \\ 18 + 6x - 12x^2 & < 0 \\ -12x^2 + 6x + 18 & < 0 \\ 12x^2 - 6x - 18 & > 0 \\ 2x^2 - x - 3 & > 0 \\ (x + 1)(2x - 3) & > 0 \end{align}
$$ x < -1 \text{ or } x > {3 \over 2} $$
\begin{align} f(x) & = x^2 (2 - x) \\ & = 2x^2 - x^3 \\ \\ f'(x) & = 2(2)x - 3x^2 \\ & = 4x - 3x^2 \\ f'(x) & > 0 \phantom{0000000000} [\text{Increasing function}] \\ 4x - 3x^2 & > 0 \\ -3x^2 + 4x & > 0 \\ 3x^2 - 4x & < 0 \\ x(3x - 4) & < 0 \end{align}
$$ 0 < x < {4 \over 3} $$
\begin{align} u & = x^2 &&& v & = 2x - 3 \\ {du \over dx} & = 2x &&& {dv \over dx} & = 2 \end{align} \begin{align} {dy \over dx} & = {(2x - 3)(2x) - (x^2)(2) \over (2x - 3)^2 } \phantom{00000} [\text{Quotient rule}] \\ & = {4x^2 - 6x - 2x^2 \over (2x - 3)^2 } \\ & = {2x^2 - 6x \over (2x - 3)^2} \\ \\ {dy \over dx} & > 0 \phantom{0000000000} [\text{Increasing function}] \\ {2x^2 - 6x \over (2x - 3)^2 } & > 0 \\ \\ \text{Since } (2x - 3)^2 > 0 & \text{ for } x > 1{1 \over 2}, \\ 2x^2 - 6x & > 0 \\ x^2 - 3x & > 0 \\ x(x - 3) & > 0 \end{align}
\begin{align} x < 0 & \text{ or } x > 3 \\ \\ \text{Since } & x > 1{1 \over 2}, \\ x & < 3 \end{align}
(i)
\begin{align} y & = 2x^3 + 3x^2 - 36x + k \\ \\ {dy \over dx} & = 2(3)x^2 + 3(2)x - 36 \\ & = 6x^2 + 6x - 36 \\ \\ {dy \over dx} & < 0 \phantom{000000000} [\text{Decreasing function}] \\ 6x^2 + 6x - 36 & < 0 \\ x^2 + x - 6 & < 0 \\ (x + 3)(x - 2) & < 0 \end{align}
$$ -3 < x < 2 $$
(ii) If the x-axis is tangent to the curve, the gradient at the point(s) is equal to 0. In addition, since the point(s) lies on the x-axis, the y-coordinates of the point(s) must be equal to 0.
\begin{align} {dy \over dx} & = 6x^2 + 6x - 36 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 6x^2 + 6x - 36 \\ 0 & = x^2 + x - 6 \\ 0 & = (x + 3)(x - 2) \end{align} \begin{align} x + 3 & = 0 & & \text{ or } & x - 2 & = 0 \\ x & = -3 & & & x & = 2 \\ \\ \therefore & \phantom{.} (-3, 0) &&& \therefore & \phantom{.} (2, 0) \\ \\ \text{Substitute } & \text{into eqn of curve,} &&& \text{Substitute } & \text{into eqn of curve,} \\ 0 & = 2(-3)^3 + 3(-3)^2 - 36(-3) + k &&& 0 & = 2(2)^3 + 3(2)^2 - 36(2) + k \\ 0 & = -54 + 27 + 108 + k &&& 0 & = 16 + 12 - 72 + k \\ 0 & = 81 + k &&& 0 & = -44 + k \\ -81 & = k &&& 44 & = k \end{align}
\begin{align} g(x) & = 2x^2 + 7x - \sqrt{6} + h(x) \\ \\ g'(x) & = 2(2)x + 7 - 0 + h'(x) \\ & = 4x + 7 + h'(x) \\ \\ \text{For } g(x) \text{ to be a } & \text{decreasing for all values of } x, \\ g'(x) < 0 \text{ } & \text{for all values of } x \\ \\ \\ \text{Let } g'(x) & = - 1 \phantom{00000} [\text{Always negative for all } x] \\ \\ -1 & = 4x + 7 + h'(x) \\ - 4x - 8 & = h'(x) \\ \\ \text{Working } & \text{backwards,} \\ h(x) & = {-4x^2 \over 2} - 8x \\ & = -2x^2 - 8x \end{align}
Question 7 - Real-life problem
(i)
\begin{align} P(x) & = 2x^3 - 21x^2 - 5 \\ \\ P'(x) & = 2(3)x^2 - 21(2)x \\ & = 6x^2 - 42x \\ \\ P'(x) & < 0 \phantom{0000000000} [\text{Profit decreasing}] \\ 6x^2 - 42x & < 0 \\ x^2 - 7x & < 0 \\ x(x - 7) & < 0 \end{align}
$$ 0 < x < 7 $$
(ii)
$P(x)$ is decreasing for $ 0 < x < 7$ and increasing from $x > 7$. Thus, this suggests that the relationship between profit and the number of doll houses is not linear.
Question 8 - Real-life problem
Note x is in hundreds
\begin{align} P(x) & = 2x^3 - 27x^2 + 108x - 5 \\ \\ P'(x) & = 2(3)x^2 - 27(2)x + 108 \\ & = 6x^2 - 54x + 108 \\ \\ P'(x) & > 0 \phantom{0000000000} [\text{Profit increasing}] \\ 6x^2 - 54x + 108 & > 0 \\ x^2 - 9x + 18 & > 0 \\ (x - 3)(x - 6) & > 0 \end{align}
\begin{align} x < 3 & \text{ or } x > 6 \\ \\ \text{Since no. of } & \text{workers } (x) > 0 , \\ 0 < x < 3 & \text{ or } x > 6 \end{align}
Profits will only increase when 1 to 299 workers or more than 600 workers are hired
(a)
B
As $x$ increases, the changes to gradient of graph: Negative → Positive → Negative → Positive → Negative →Positive
(b)
C
As $x$ increases, the changes to gradient of graph: Positive → Negative → Positive
(c)
A
As $x$ increases, the changes to gradient of graph: Positive → Negative → Negative → Positive