V(t) = 30 + 12t² - t³, 0 ≤ t ≤ 8
Given t = 4, dt = 4\(\frac16\) - 4 = \(\frac16\)
Approximate change in voters = (24t -3t²)dt
= [24(4) - 3(4)²] = \(\frac16\)
= (96 - 48)\(\frac16\) = \(\frac{48}{6}\) = 8
Since the function is given in thousands approximate change in voters = 8000

log₁₀e = 0.4343 to find log₁₀g 1003
f(1000) = log₁₀1000 = log₁₀10³ = 3log₁₀10³ = 3 log₁₀10
= 3(1) = 3
f'(x) = \(\frac1x\). log10^e
f'(1000) = \(\frac{1}{1000}\)(0.4343)
∴ L(x) = f(x₀) f'(x₀) (x - x₀)
= 3 + \(\frac{1}{1000}\) (0.4343) (3)
= 3 + \(\frac{1.3029}{1000}\)
= 3 + 0.0013029
log₁₀1003 = 3.0013029

Given f(x) = x³ - 2x² ; x = 2, ∆ x = dx = 0.5
(1) df = f'(x).∆x = (3.x² - 4x) ∆x
= [3(2)2 - 4(2)] (0.5)
= 4(0.5) = 2.0
∆f = f(x + ∆x) - f(x)
= f(2.5) - f(2)
= [(2.5)³ - 2 (2.5)²] - [2³ - 2(2²)]
= 15.625 -12.5 -0 = 3.125
(2) df = f'(x) ∆x (2x + 2) (∆x)
x = -0.5, ∆x = dx = 0.1
= (2(-0.05) + 2) = 0.1
∆f = f(x + ∆x) - f(x)
= f(-0.5 + 0.1) - f(-0.5)
= f(0.4) - f(-0.5)
= [(-0.4)² + 2(-0.4) + 3] - [(-0.5)² + 2(-0.5) + 3]
= (16 - 0.8 + 3) - (0.25 - 1 + 3)
= 2.36 - 2.25 = 0.11 

Actual value = 12.5 cm,
Approximate value = 12.65 cm
Area of the circular plate = πr²
Percentage error = 0.024 x 100 = 2.4%
Volume of sphere = \(\frac43\)πr² 

Actual radius of the circular plate = 12.5 cm
Measured radius of the circular plate = 12.65
dr = 12.65-12.5
= 0.15
\( \mathrm{A} =\pi \mathrm{r}^{2} \\ \mathrm{dA} =2 \pi \mathrm{rdv} \)
Change in Area
A(12.65)-A(12.5) = dA
\( =2 \pi \times 12.5 \times 0.15 \)
\(=3.75 \pi \)
Absolute error = 3.7725\(\pi\) - 3.75\(\pi\)
:0.0225\(\pi\) cm²

\({ h }({ x }_{ o })=\frac { x }{ 1+1 } =\frac { 1 }{ 2 } \)
\({ h }^{ ' }(x)=\frac { (x+1)(1)-x(1) }{ { (x+1) }^{ 2 } } \)
\(\frac { x+1-x }{ { (x+1) }^{ 2 } } =\frac { 1 }{ ({ x+1) }^{ 2 } } \)
\({ h }^{ ' }({ x }_{ o })=\frac { 1 }{ { 2 }^{ 2 } } =\frac { 1 }{ 4 } \)
∴ L(x) = h(x_o) + h'(x₀)(x - x_o)
\(=\frac { 1 }{ 2 } +\frac { 1 }{ 4 } (x-1)=\frac { 2+x-1 }{ 4 } =\frac { x+1 }{ 4 } \)
∴ L(x) = \(\frac { x+1 }{ 4 } \)

Given \(g(x)=\sqrt { { x }^{ 2 }+9 } ,{ x }_{ 0 }=-4\)
\(g(x)=\sqrt { { (-4) }^{ 2 }+9 } =\sqrt { 16+9 } =5\)
\({ g }^{ ' }(x)=\frac { 1 }{ 2 } ({ { x }^{ 2 }+9 })^{ -\frac { 1 }{ 2 } }(2x)=\frac { x }{ \sqrt { { x }^{ 2 }+9 } } \)
\(\therefore { g }^{ ' }({ x }_{ 0 })=\frac { -4 }{ \sqrt { { (-4) }^{ 2 }+9 } } =\frac { -4 }{ 5 } \)
∴ L(x) = g(x_o) + g'(x₀)(x - x_o)
= \(5-\frac { 4 }{ 5 } (x+4)=\frac { 25-4x-16 }{ 5 } \)
L(x) = \(\frac { 9-4x }{ 5 } \)

y = 52 \(\sqrt { x } \), 0, ≤ x ≤ 9
Give x = 1, dx = 1.1-1 = 0.1
Approximate number of words learned
= \(52.\frac { 1 }{ 2 } { x }^{ \frac { 1 }{ 2 } -1 }\) = \(\frac { 26 }{ \sqrt { x } } \) dx
= \(\frac { 26 }{ \sqrt { 1 } } \) (0.1) = 2.6
≃ 3 words
(ii) Then approximate number of words learned
= \(\frac { 26 }{ \sqrt { x } } \)dx = \(\frac { 26 }{ \sqrt { 4} } \)(0.1) = \(\frac{26}{4}\)(0.1)
= 13 (0.1) = 1.3
≃ 1 word

When x = 4, dx = 4.1-4 = 0.1