Detailed Answer

a)

\[ (4 - 8x)^2 = (4)^2 - 2 \times 4 \times 8x + (8x)^2 = 16 - 64x + 64x^2 \]

This confirms the given expression.

b) To find the value of \(\int_{0}^{3} (4 - 8x)^2 \, dx\), we use the result from part (a), where:

\[ (4 - 8x)^2 = 16 - 64x + 64x^2 \]

Thus, the integral becomes:

\[ \int_{0}^{3} (16 - 64x + 64x^2) \, dx \]

\[ \int_{0}^{3} (16 - 64x + 64x^2) = 16x - 32x^2 + \frac{64}{3}x^3 \, dx \]

\[ (16 - 32(3)^2 + \frac{64}{3}3^3) - 16 = 336 \]

Detailed Answer

a) To find the equation of the tangent to \( f(x) \) at point \( Q(2, 14) \), we first need to determine the gradient of the function at \( x = 2 \).

\[ f'(2) = 9(2)^2 - 4(2) + 1 \]

\[ f'(2) = 36 - 8 + 1 = 29 \]

So, the gradient of the tangent at \( x = 2 \) is 29.

Using the point-slope form of the equation of the tangent line:

\[ y - 14 = 29(x - 2) \]

\[ y - 14 = 29x - 58 \]

\[ y = 29x - 44 \]

Thus, the equation of the tangent line at point \( Q \) is \( y = 29x - 44 \).

b) To find the equation of \( f(x) \), we integrate \( f'(x) \).

\[ f(x) = 3x^3 - 2x^2 + x + C \]

To find the constant \( C \), use the point \( Q(2, 14) \):

\[ 14 = 3(2)^3 - 2(2)^2 + 2 + C \]

\[ C = 14 - 18 = -4 \]

So, the equation of \( f(x) \) is:

\[ f(x) = 3x^3 - 2x^2 + x - 4 \]

Detailed Answer

a)

\[ a = \int_{2}^{3} (2x^2 - 3x - 2) \, dx \]

b)

\[ \int (2x^2 - 3x - 2) \, dx = \frac{2x^3}{3} - \frac{3x^2}{2} - 2x \]

Evaluating this from \( x = 2 \) to \( x = 3 \):

Substitute \( x = 3 \):

\[ \frac{2(3)^3}{3} - \frac{3(3)^2}{2} - 2(3) = \frac{54}{3} - \frac{27}{2} - 6 \]

\[ = 18 - 13.5 - 6 = -1.5 \]

Substitute \( x = 2 \):

\[ \frac{2(2)^3}{3} - \frac{3(2)^2}{2} - 2(2) = \frac{16}{3} - \frac{12}{2} - 4 \]

\[ = \frac{16}{3} - 6 - 4 = \frac{16}{3} - 10 = -\frac{14}{3} \]

The area \( a \) is:

\[ a = \left( -1.5 \right) - \left( -\frac{14}{3} \right) = -1.5 + \frac{14}{3} = \frac{19}{6} \]

So the area \( a \) is:

\[ a = \frac{19}{6} \]

c) To find the value of \( c \), which is measured as \( c = \frac{b}{2a} \), we first need to find \( b \). To do that, we need to find the top side of the rectangle which it forms, which can be done by finding \( f(3) \)

\[ f(3) = 7 * 1 = 7 \]

So the graph of this function looks as follows:

So, we have that:

\[ b = 7 - \frac{19}{6} = \frac{23}{6} \]

So, finally we have that

\[ c = \frac{b}{2a} = \frac{\frac{23}{6}}{\frac{19}{6}} = \frac{23}{19} \]

Detailed Answer

a) To find the equation of \( f(x) \), we integrate the given gradient function \( f'(x) = kx + 5 \):

\[f'(x) = kx + 5\]

Integrating \( f'(x) \) with respect to \( x \):

\[f(x) = \int (kx + 5) \, dx = \frac{kx^2}{2} + 5x + C\]

Given that \( f(x) \) cuts the y-axis at \( y = 2 \), we know \( f(0) = 2 \):

\[f(0) = \frac{k \cdot 0^2}{2} + 5 \cdot 0 + C = 2 \implies C = 2\]

So, the equation becomes:

\[f(x) = \frac{kx^2}{2} + 5x + 2\]

Given that \( f(x) \) cuts the x-axis at \( x = 2 \), we know \( f(2) = 0 \):

\[f(2) = \frac{k \cdot 2^2}{2} + 5 \cdot 2 + 2 = 0 \implies 2k + 10 + 2 = 0 \implies 2k + 12 = 0 \implies k = -6\]

Thus, the function \( f(x) \) is:

\[f(x) = \frac{-6x^2}{2} + 5x + 2 = -3x^2 + 5x + 2\]

b) This can be easily found in the GDC:

\[\text{Area} = \int_{0}^{2} (-3x^2 + 5x + 2) \, dx = 6\]

c) To find the tangent at \( x = 1 \), we need the slope at \( x = 1 \). The slope is given by \( f'(x) \):

\[f'(x) = kx + 5 = -6x + 5\]

Evaluating the slope at \( x = 1 \):

\[f'(1) = -6 \cdot 1 + 5 = -1\]

The slope of the tangent line at \( x = 1 \) is \( -1 \).

Next, we find the y-coordinate at \( x = 1 \):

\[f(1) = -3(1)^2 + 5 \cdot 1 + 2 = -3 + 5 + 2 = 4\]

So the point of tangency is \( (1, 4) \).

Using the point-slope form of the equation of a line:

\[y - y_1 = m(x - x_1)\]

Where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency:

\[y - 4 = -1(x - 1)\]

Simplifying:

\[y - 4 = -x + 1 \implies y = -x + 5\]

Detailed Answer

a) To find the equation of the tangent to \( f(x) = -2x^3 + 6x^2 \) at \( x = 3 \) we have to first find the derivative:

\[ f(x) = -2x^3 + 6x^2 \]

\[ f'(x) = \frac{d}{dx}(-2x^3 + 6x^2) = -6x^2 + 12x \]

\[ f'(3) = -6(3)^2 + 12(3) = -54 + 36 = -18 \]

So, the slope of the tangent at \( x = 3 \) is \(-18\).

Now, we find the y-coordinate at \( x = 3 \):

\[ f(3) = -2(3)^3 + 6(3)^2 = -54 + 54 = 0 \]

So, the point of tangency is \( (3, 0) \).

Then, we find the tangent line:

\[ y - y_1 = m(x - x_1) \]

Substituting \( m = -18 \), \( x_1 = 3 \), and \( y_1 = 0 \):

\[ y - 0 = -18(x - 3) \implies y = -18x + 54 \]

b) To find the shaded area we first have to find the x-intercepts:

\[ f(x) = -2x^3 + 6x^2 = x^2(-2x + 6) \]

\[ x^2(-2x + 6) = 0\]

\[ x = 0, x = 3\]

Now, we can plug the function into the GDC and find the area:

So, the area is equal to 13.5.

c)

\[ \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} \]

Given the vertices \( A(1, 0) \), \( B(5, 0) \), and \( C(3, y) \), the base \( AB \) is \( 5 - 1 = 4 \).

\[ \frac{1}{2} \times 4 \times y = 13.5 \]

\[ 2y = 13.5 \implies y = \frac{13.5}{2} = 6.75 \]

The y-coordinate of point \( C \) is \( 6.75 \).

Detailed Answer

a)

\[ \frac{4\sqrt{z} + 20}{\sqrt{z}} = \frac{4\sqrt{z}}{\sqrt{z}} + \frac{20}{\sqrt{z}} \]

We know that \(\frac{4\sqrt{z}}{\sqrt{z}} = 4\) and \(\frac{20}{\sqrt{z}} = 20z^{-\frac{1}{2}}\). Therefore, the expression becomes:

\[ 4 + 20z^{-\frac{1}{2}} \]

This matches the form \( 4 + 20z^p \) where \( p = -\frac{1}{2} \).

b) To find the value of \( \int_{1}^{16} \frac{4\sqrt{z} + 20}{\sqrt{z}} \, dz \), we use the simplified form \( 4 + 20z^{-\frac{1}{2}} \):

\[ \int_{1}^{16} \left( 4 + 20z^{-\frac{1}{2}} \right) \, dz \]

\[ \int_{1}^{16} \left( 4 + 20z^{-\frac{1}{2}} \right) \, dz = \int_{1}^{16} 4 \, dz + \int_{1}^{16} 20z^{-\frac{1}{2}} \, dz \]

Evaluating the integrals:

\[ \int_{1}^{16} 4 \, dz = 4z \bigg|_{1}^{16} = 4(16) - 4(1) = 64 - 4 = 60 \]

\[ \int_{1}^{16} 20z^{-\frac{1}{2}} \, dz = 20 \left( 2z^{\frac{1}{2}} \right) \bigg|_{1}^{16} = 40z^{\frac{1}{2}} \bigg|_{1}^{16} \]

Evaluating this at the bounds:

\[ 40z^{\frac{1}{2}} \bigg|_{1}^{16} = 40(16^{\frac{1}{2}}) - 40(1^{\frac{1}{2}}) = 40(4) - 40(1) = 160 - 40 = 120 \]

Adding the results of both integrals:

\[ 60 + 120 = 180 \]

Thus, the value of the integral is \( 180 \).

Detailed Answer

a) As it can be seen, we have \( -x^2 \) for \( f(x) \), meaning that i's concave down, so it will be green. For \( g(x) \) we have \( x^2 \), so it is concave up, which means that it is the red curve.

b) To find the points of intersection of the functions, we set \( f(x) = g(x) \):

\[ -x^2 + 3x - 11 = x^2 - 5x - 5 \]

\[ -x^2 + 3x - 11 - x^2 + 5x + 5 = 0 \]

\[ -2x^2 + 8x - 6 = 0 \]

\[ x^2 - 4x + 3 = 0 \]

\[ (x - 1)(x - 3) = 0 \]

So, the solutions are:

\[ x = 1 \quad \text{and} \quad x = 3 \]

Since \( a > b \):

\[ a = 3 \quad \text{and} \quad b = 1 \]

c) To find the area enclosed by the two functions between \( x = 1 \) and \( x = 3 \), we integrate the difference \( f(x) - g(x) \) from 1 to 3. We start with \( f(x) \) as this is the function on top:

\[ \text{Area} = \int_{1}^{3} [f(x) - g(x)] \, dx \]

Substitute \( f(x) \) and \( g(x) \):

\[ f(x) = -x^2 + 3x - 11 \]

\[ g(x) = x^2 - 5x - 5 \]

\[ f(x) - g(x) = (-x^2 + 3x - 11) - (x^2 - 5x - 5) \]

\[ f(x) - g(x) = -2x^2 + 8x - 6 \]

Now, integrate \( -2x^2 + 8x - 6 \) from 1 to 3:

\[ \int_{1}^{3} (-2x^2 + 8x - 6) \, dx \]

\[ \int (-2x^2 + 8x - 6) \, dx = -\frac{2x^3}{3} + 4x^2 - 6x \]

Evaluate this from 1 to 3:

\[ \left[ -\frac{2x^3}{3} + 4x^2 - 6x \right]_{1}^{3} \]

First, evaluate at \( x = 3 \):

\[ -\frac{2(3)^3}{3} + 4(3)^2 - 6(3) = -\frac{54}{3} + 36 - 18 = -18 + 36 - 18 = 0 \]

Next, evaluate at \( x = 1 \):

\[ -\frac{2(1)^3}{3} + 4(1)^2 - 6(1) = -\frac{2}{3} + 4 - 6 = -\frac{2}{3} - 2 = -\frac{2 + 6}{3} = -\frac{8}{3} \]

The area is:

\[ 0 - \left( -\frac{8}{3} \right) = \frac{8}{3} \]

Thus, the area enclosed by the two functions is \( \frac{8}{3} \) square units.

Detailed Answer

To solve the integral, we use the trigonometric identity:

\[ \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \]

Applying this identity with \( \theta = \frac{x}{4} \):

\[ \cos^2\left(\frac{x}{4}\right) = \frac{1 + \cos\left(\frac{x}{2}\right)}{2} \]

Substitute this into the integral:

\[ \int_{0}^{2\pi} \cos^2\left(\frac{x}{4}\right) \, dx = \int_{0}^{2\pi} \frac{1 + \cos\left(\frac{x}{2}\right)}{2} \, dx \]

Split this into two separate integrals:

\[ \int_{0}^{2\pi} \frac{1 + \cos\left(\frac{x}{2}\right)}{2} \, dx = \frac{1}{2} \int_{0}^{2\pi} 1 \, dx + \frac{1}{2} \int_{0}^{2\pi} \cos\left(\frac{x}{2}\right) \, dx \]

Evaluate each integral separately:

1. Integral of 1:

\[ \frac{1}{2} \int_{0}^{2\pi} 1 \, dx = \frac{1}{2} \left[x\right]_{0}^{2\pi} = \frac{1}{2} \cdot (2\pi - 0) = \pi \]

2. Integral of \( \cos\left(\frac{x}{2}\right) \):

Use substitution \( u = \frac{x}{2} \), hence \( du = \frac{1}{2} dx \) or \( dx = 2 \, du \). Adjust the limits:

When \( x = 0 \), \( u = 0 \). When \( x = 2\pi \), \( u = \pi \).

\[ \frac{1}{2} \int_{0}^{2\pi} \cos\left(\frac{x}{2}\right) \, dx = \frac{1}{2} \int_{0}^{\pi} \cos(u) \cdot 2 \, du = \int_{0}^{\pi} \cos(u) \, du \]

Integrate:

\[ \int_{0}^{\pi} \cos(u) \, du = \sin(u) \Big|_{0}^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0 \]

Combine the results:

\[ \frac{1}{2} \int_{0}^{2\pi} 1 \, dx + \frac{1}{2} \int_{0}^{2\pi} \cos\left(\frac{x}{2}\right) \, dx = \pi + 0 = \pi \]

Thus, the value of the integral is:

\[ \int_{0}^{2\pi} \cos^2\left(\frac{x}{4}\right) \, dx = \pi \]

Detailed Answer

To find the function \( f(x) \), we need to integrate \( f'(x) \):

\[ f(x) = \int \frac{5x}{7x^2 + 3} \, dx \]

Substitute \( u = 7x^2 + 3 \):

\[ du = 14x \, dx \text{ or } dx = \frac{du}{14x} \]

Substitute into the integral:

\[ \int \frac{5x}{7x^2 + 3} \, dx = \int \frac{5x}{u} \times \frac{du}{14x} \]

\[ = \frac{5}{14} \int \frac{du}{u} \]

\[ = \frac{5}{14} \ln|u| + C \]

Substitute back \( u = 7x^2 + 3 \):

\[ f(x) = \frac{5}{14} \ln|7x^2 + 3| + C \]

To find \( C \), use the point \( Q(1, 3) \):

\[ 3 = \frac{5}{14} \ln|7 \times 1^2 + 3| + C \]

\[ 3 = \frac{5}{14} \ln|10| + C \]

Solving for \( C \):

\[ C = 3 - \frac{5}{14} \ln 10 \]

The final equation for \( f(x) \) is:

\[ f(x) = \frac{5}{14} \ln \left|\frac{7x^2 + 3}{10}\right| + 3 \]

Detailed Answer

a)

\[ f(x) = -4\cos^2(x) - 2\sin^2(x) + 2 \]

Rewrite \( f(x) \) using the identity \( \sin^2(x) = 1 - \cos^2(x) \):

\[ f(x) = -4\cos^2(x) - 2(1 - \cos^2(x)) + 2 \]

\[ f(x) = -4\cos^2(x) - 2 + 2\cos^2(x) + 2 \]

\[ f(x) = -2\cos^2(x) \]

Set \( f(x) = 0 \):

\[ -2\cos^2(x) = 0 \]

\[ \cos^2(x) = 0 \]

\[ \cos(x) = 0 \]

The cosine function is zero at \( x = \frac{\pi}{2} \) within the interval \( 0 \leq x \leq \pi \). Therefore, the root is:

\[ x = \frac{\pi}{2} \]

b)

\[ f(x) = -2\cos^2(x) \]

Using the chain rule:

\[ \frac{d}{dx}[\cos^2(x)] = 2\cos(x)(-\sin(x)) = -2\cos(x)\sin(x) \]

Thus:

\[ f'(x) = -2 \times (-2\cos(x)\sin(x)) = 4\cos(x)\sin(x) \]

\[ f'(x) = 2\sin(2x) \]

So, \( a = 2 \) and \( b = 2 \).

c)

\[ 2\sin(2x) = 0 \]

\[ \sin(2x) = 0 \]

The general solution for \( \sin(2x) = 0 \) is:

\[ 2x = n\pi \text{ for integer } n \]

\[ x = \frac{n\pi}{2} \]

Within the interval \( 0 \leq x \leq \pi \), the values are \( x = 0 \) and \( x = \frac{\pi}{2} \).

We substitute these x-values into \( f(x) \) to find the corresponding y-values:

For \( x = 0 \):

\[ f(0) = -2\cos^2(0) = -2 \cdot 1^2 = -2 \]

For \( x = \frac{\pi}{2} \):

\[ f\left(\frac{\pi}{2}\right) = -2\cos^2\left(\frac{\pi}{2}\right) = -2 \cdot 0^2 = 0 \]

So, the coordinates are: \( \left(0, -2\right), \ \left(\frac{\pi}{2}, 0\right) \)

d) We use what we found previously:

\[ -4\cos^2(x) - 2\sin^2(x) + 2= -2\cos^2(x) \]

Rewrite \( \cos^2(x) \) using \( \cos^2(x) = \frac{1 + \cos(2x)}{2} \):

\[ \int_{0}^{\pi} -2 \left(\frac{1 + \cos(2x)}{2}\right) \, dx = \int_{0}^{\pi} -1 - \cos(2x) \, dx \]

Integrate term by term:

\[ \int_{0}^{\pi} -1 \, dx - \int_{0}^{\pi} \cos(2x) \, dx \]

Using the substitution \( u = 2x \), \( du = 2 \, dx \), so \( dx = \frac{du}{2} \):

\[ \int_{0}^{\pi} \cos(2x) \, dx = \frac{1}{2} \int_{0}^{2\pi} \cos(u) \, du \]

\[ \int \cos(u) \, du = \sin(u) \]

\[ \frac{1}{2} [\sin(u)]_{0}^{2\pi} = \frac{1}{2} [\sin(2\pi) - \sin(0)] = 0 \]

So:

\[ \int_{0}^{\pi} -1 \, dx = -x \Big|_{0}^{\pi} = -\pi \]

\[ \int_{0}^{\pi} \cos(2x) \, dx = 0 \]

Thus:

\[ \int_{0}^{\pi} -2\cos^2(x) \, dx = -\pi \]

Detailed Answer

a) To find the inverse function \( g^{-1}(x) \), we start with the function \( g(x) = \sqrt{3x + 4} \). Set \( y = \sqrt{3x + 4} \) and solve for \( x \):

\[ y = \sqrt{3x + 4} \]

Square both sides to eliminate the square root:

\[ y^2 = 3x + 4 \]

Rearrange to solve for \( x \):

\[ 3x = y^2 - 4 \]

\[ x = \frac{y^2 - 4}{3} \]

Switch \( x \) and \( y \) to find the inverse function:

\[ g^{-1}(x) = \frac{x^2 - 4}{3} \]

b) To find the point of intersection we can plug both functions into the GDC:

The coordinates of intersection are \( (4, 4) \).

c)

\[ g'(x) = \frac{1}{2}(3x + 4)^{-1/2} \times 3 = \frac{3}{2\sqrt{3x + 4}} \]

d) We first find the derivative of \( g^{-1}(x) \):

\[ g^{-1}(x) = \frac{x^2 - 4}{3} \]

\[ g^{-1}(x)' = \frac{2x}{3} \]

Equalize both derivatives:

\[ \frac{3}{2\sqrt{3x + 4}} = \frac{2x}{3} \]

Now we can plug them both into the GDC as well:

So, the gradients are the same for \( x = 0.874 \)

e) The area enclosed will look like the one presented on the graph below:

To find this area, we calculate the following integral. This can be easily done in the GDC:

\[ \int_{0}^{4} \left( \sqrt{3x + 4} - \frac{x^2 - 4}{3} \right) \, dx \]