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:

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