Key Concept: Sum of Squares

b) $$x = 10, y = 10$$
[Solution Description]
Let the numbers be x and y. The constraint is $$x + y = 20$$. We want to minimize $$S = x^2 + y^2$$.

Substitute $$y = 20 – x$$ into the equation for S:

$$
S = x^2 + (20-x)^2
$$

Simplifying:

$$
S = x^2 + 400 – 40x + x^2 = 2x^2 – 40x + 400
$$

Take the derivative with respect to x:

$$
\frac{dS}{dx} = 4x – 40
$$

Set the derivative equal to zero to find critical points:

$$
4x – 40 = 0 \Rightarrow x = 10
$$

Check the second derivative to confirm it’s a minimum:

$$
\frac{d^2S}{dx^2} = 4 > 0
$$

Since the second derivative is positive, x = 10 is a minimum point. Therefore, the numbers are x = 10 and y = 10.

Key Concept: Rate of Change in Geometry

c) $$2\pi \, \text{cm/s}$$
[Solution Description] The circumference C of a circle is given by $$C = 2\pi r$$. The rate of change of the circumference with respect to time t is determined by differentiating:$$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$$.
Given that $$\frac{dr}{dt} = 1 \text{ cm/s}$$,$$\frac{dC}{dt} = 2\pi \times 1 = 2\pi \text{ cm/s}$$

Key Concept: Advanced First Derivative Test, Inflection and Concavity

a) Maxima at $$x = 1$$; Minima at $$x = -1$$; Inflection at $$x = \pm \sqrt{2}$$

[Solution Description]

To find the critical points, we first compute the derivative:

$$f'(x) = 5x^4 – 15x^2 + 10$$

Set $$f'(x) = 0$$ to find critical points:

$$5x^4 – 15x^2 + 10 = 0$$

Simplifying, divide everything by 5:

$$x^4 – 3x^2 + 2 = 0$$

Letting $$u = x^2$$, this becomes:

$$u^2 – 3u + 2 = 0$$

Factor the quadratic:

$$(u-1)(u-2) = 0$$

So, $$u = 1$$ or $$u = 2$$. Converting back to $$x^2$$ gives $$x^2 = 1$$ or $$x^2 = 2$$. Hence, $$x = \pm 1$$ or $$x = \pm \sqrt{2}$$.

Use the first derivative test around these points to classify them. Calculating second derivatives also helps:

$$f”(x) = 20x^3 – 30x$$

Evaluate $$f”$$ at each critical point:

– At $$x = \pm 1$$: $$f”(\pm 1) = 20(\pm 1)^3 – 30(\pm 1) = \mp 10$$, negative for $$x = 1$$ indicating local maximum, positive for $$x = -1$$ indicating local minimum.

– At $$x = \pm \sqrt{2}$$: Check that $$f”(\pm \sqrt{2})$$ changes signs indicating inflection points.

We conclude:

– Local maxima at $$x = 1$$

– Local minima at $$x = -1$$

– Inflection points at $$x = \pm \sqrt{2}$$

Key Concept: Simple Application

c) Local minimum at x = 2
[Solution Description] First, find the derivative: $$f'(x) = 2x – 4$$. Set it to zero: $$2x – 4 = 0 \Rightarrow x = 2$$. Check the second derivative: $$f”(x) = 2 > 0$$, indicating a local minimum at x = 2.

Key Concept: Revenue Function Analysis

a) 96
[Solution Description]
To find the marginal revenue, we differentiate the revenue function R(x) with respect to x, giving us:

$$\frac{dR}{dx} = MR = 6x + 36$$

Substitute x = 10 into the marginal revenue equation:$$MR = 6(10) + 36 = 60 + 36 = 96$$
Thus, the marginal revenue at x = 10 is 96.

Key Concept: Marginal Revenue

b) 80
[Solution Description]
Marginal revenue is determined by differentiating R(x) with respect to x: $$\frac{dR}{dx}$$.
$$
\frac{dR}{dx} = 10x + 40
$$
Evaluate at x = 4:
$$
MR = 10(4) + 40
= 40 + 40
= 80
$$

Key Concept: Increasing/Decreasing Functions

a) $$f'(x) > 0$$
[Solution Description] A function f(x) is said to be increasing on an interval if its first derivative $$f'(x)$$ is greater than zero $$(f'(x) > 0)$$ for all points in that interval. This means that as x increases, f(x) also increases.

Key Concept: Slope Calculation

b) $$f'(a)$$
[Solution Description] The slope of the tangent line to the curve y = f(x) at a particular point x = a is given by the derivative evaluated at that point, i.e., $$f'(a)$$.

Key Concept: Function Continuity

a) Yes
[Solution Description] To check continuity in the interval (1, 2), first simplify the function:

$$\frac{x^2 – 1}{x – 1} = \frac{(x – 1)(x + 1)}{x – 1}$$.

For all x except x = 1, this expression simplifies to x + 1.

Since the interval (1, 2) does not include x = 1, the function is the same as x + 1 in this region, hence it is continuous in the interval.

Key Concept: Approximation Error

a) When the function is linear
[Solution Description] The error in linear approximation is minimized when the function being approximated is linear because the linear approximation will exactly match the function.

Key Concept: Inflection Point Complexity, Optimization Challenge

a) Each side $$\frac{L}{4}$$ (square)
[Solution Description] Let the sides be x and y, thus $$2x + 2y = L$$. Maximize A = xy.

Express y in terms of x:
$$
y = \frac{L}{2} – x
$$

Substitute into A = xy:
$$
A = x\left(\frac{L}{2} – x\right) = \frac{Lx}{2} – x^2
$$

Differentiate A w.r.t. x:
$$
\frac{dA}{dx} = \frac{L}{2} – 2x
$$

Set $$\frac{dA}{dx} = 0$$ for optimum:
$$
\frac{L}{2} – 2x = 0 \rightarrow x = \frac{L}{4}
$$

This gives $$y = \frac{L}{4}$$ as well, showing square configuration maximizes area. Verify $$A” < 0$$, confirming a maximum by concavity check. Therefore, sides are equal, each $$\frac{L}{4}$$, maximizing enclosed area.

Key Concept: Chain Rule Application, Complex Function Rate

a) $$-\frac{1}{8}$$

[Solution Description]

To find $$\frac{dy}{dx}$$, we need to use the chain rule:

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$

First, calculate $$\frac{dy}{dt}$$:

$$y(t) = t^3 – 4t$$

$$\frac{dy}{dt} = 3t^2 – 4$$

Now, calculate $$\frac{dx}{dt}$$:

$$x(t) = 3t^2 + 2t$$

$$\frac{dx}{dt} = 6t + 2$$

At $$t = 1$$, substitute the values:

$$\frac{dy}{dt} = 3(1)^2 – 4 = 3 – 4 = -1$$

$$\frac{dx}{dt} = 6(1) + 2 = 6 + 2 = 8$$

Therefore:

$$\frac{dy}{dx} = \frac{-1}{8} = -\frac{1}{8}$$

Key Concept: Simple Derivatives

a) $$\frac{2x}{x^2 + 1}$$
[Solution Description]
To find the derivative of $$ f(x) = \ln(x^2 + 1) $$, apply the chain rule.

Step 1: Identify the outer function and its derivative:
– Outer function: $$\ln(u)$$, where $$u = x^2 + 1$$.
– Derivative of $$\ln(u)$$ is:
$$
\frac{1}{u}
$$

Step 2: Take the derivative of the inner function:
– Inner function: $$u = x^2 + 1$$.
– Derivative of u is:
$$
\frac{d}{dx}(x^2+1) = 2x
$$

Step 3: Apply chain rule:
$$
\frac{df}{dx} = \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1}
$$

Therefore, the derivative is $$\frac{2x}{x^2 + 1}$$.

Key Concept: Critical Points Analysis, Interval Analysis

c) Increase: $$(3, \infty)$$; Decrease: $$(0, 3)$$; Local Max: $$x = 0$$

[Solution Description]

First, find the derivative $$f'(x)$$:

$$f'(x) = 4x^3 – 24x^2 + 36x = 4x(x^2 – 6x + 9)$$

Factor further using $$(x-3)^2$$:

$$f'(x) = 4x(x-3)^2$$

Critical points occur when $$f'(x) = 0$$, giving $$x = 0$$ and $$x = 3$$.

Use the first derivative test to determine behavior around these points:

For $$x < 0$$: $$f'(x) > 0$$

At $$x = 0$$: change from positive to negative, indicating a local maximum.

For $$0 < x < 3$$: $$f'(x) < 0$$ At $$x = 3$$: no change in sign, indicating an inflection point. For $$x > 3$$: $$f'(x) > 0$$

Thus, the interval of increase is $$(3, \infty)$$, the interval of decrease is $$(0, 3)$$, and there’s a local maximum at $$x=0$$.

Key Concept: Balloon Radius

b) $$\frac{5}{4\pi}\text{ cm/s}$$
[Solution Description] Volume $$V = \frac{4}{3}\pi r^3$$. Differentiate with respect to t:
$$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$$
Given $$\frac{dV}{dt} = 500$$, solve for $$\frac{dr}{dt}$$ when r = 10:
$$500 = 4\pi (10)^2 \frac{dr}{dt} \rightarrow 500 = 400\pi \frac{dr}{dt} \rightarrow \frac{dr}{dt} = \frac{500}{400\pi} = \frac{5}{4\pi} \text{ cm/s}$$

Key Concept: Basic Tangent Equation

a) $$ y = 3x – 2 $$
[Solution Description]
To find the equation of the tangent line, we first calculate the slope by differentiating $$ y = x^3 $$. The derivative is:
$$
f'(x) = \frac{d}{dx}(x^3) = 3x^2.
$$

Evaluating this at $$ x = 1 $$:
$$
f'(1) = 3(1)^2 = 3.
$$

Now, using the point-slope form of the equation of a line:
$$
y – 1 = 3(x – 1).
$$

Simplifying gives:
$$
y = 3x – 2.
$$

Key Concept: Local Extrema

a) Maxima at x=1, Minima at x=3
[Solution Description] To find the local maxima and minima, we first find the derivative of the function:
$$\frac{d}{dx}(x^3 – 6x^2 + 9x + 1) = 3x^2 – 12x + 9$$.

Set this equal to zero to find critical points:
$$
3x^2 – 12x + 9 = 0
$$

Divide through by 3:
$$
x^2 – 4x + 3 = 0
$$

Factor the quadratic equation:
$$(x-1)(x-3)=0$$.

So, the critical points are x=1 and x=3.

Next, apply the second derivative test. The second derivative is:
$$\frac{d^2}{dx^2}(x^3 – 6x^2 + 9x + 1) = 6x – 12$$.

Evaluate at x=1: $$6(1) – 12 = -6$$, which is less than 0, so there is a local maximum at x=1.

Evaluate at x=3: $$6(3) – 12 = 6$$, which is greater than 0, so there is a local minimum at x=3.

Therefore, the answer is Maxima at x=1, Minima at x=3.

Key Concept: Simple Derivative

c) $$12$$
[Solution Description] We need to differentiate $$y = x^3$$ with respect to x to find $$\frac{dy}{dx}$$:
$$
\frac{dy}{dx} = \frac{d}{dx}(x^3) = 3x^2
$$ Evaluating at x = 2 gives:
$$
\frac{dy}{dx} = 3 \times 2^2 = 3 \times 4 = 12
$$

Key Concept: Increasing or Decreasing Intervals

a) Increasing on $$(-\infty, 1) \quad \text{and} \quad (3, \infty)$$
[Solution Description]
Find the first derivative of g(x):

$$
g'(x) = 3x^2 – 12x + 9.
$$

Solve $$g'(x) > 0$$ to find increasing intervals. Factor $$g'(x)$$:

$$
3x^2 – 12x + 9 = 3(x^2 – 4x + 3) = 3(x-1)(x-3).
$$

The critical points are x = 1 and x = 3. Test intervals around these points:

– For $$x < 1$$ (choose x = 0): $$g'(0) = 3(0-1)(0-3) = 9 > 0$$.
– For $$1 < x < 3$$ (choose x = 2): $$g'(2) = 3(2-1)(2-3) = -3 < 0$$. - For $$x > 3$$ (choose x = 4): $$g'(4) = 3(4-1)(4-3) = 9 > 0$$.

Therefore, g(x) is Increasing on $$(-\infty, 1) \quad \text{and} \quad (3, \infty)$$

Key Concept: Nearest Point

b) $$(3, \sqrt{8})$$
[Solution Description]
To determine the nearest point on the hyperbola to (3, 0), parameterize the hyperbola as $$(\cosh t, \sinh t)$$. The distance squared from (3, 0) to $$(\cosh t, \sinh t)$$ is:
$$
D(t) = (\cosh t – 3)^2 + \sinh^2 t.
$$
Expanding,
$$
D(t) = \cosh^2 t – 6 \cosh t + 9 + \sinh^2 t = 1 + 2\sinh^2 t – 6 \cosh t + 9.
$$
Minimize D(t) using first derivative test. After solving, the minimized distance yields the closest point on hyperbola as approximately $$(3, \pm \sqrt{8})$$.

Key Concept: Simple Maxima

c) 50
[Solution Description]

Let the width of the rectangle be x. Then its height y can be expressed as the function of x:

$$
y = \sqrt{100 – x^2}
$$

The area A of the rectangle is:

$$
A = xy = x\sqrt{100 – x^2}
$$

To maximize A, take the derivative of A with respect to x and set it equal to zero:

$$
\frac{dA}{dx} = \sqrt{100-x^2} + \left(\frac{-x^2}{\sqrt{100-x^2}}\right)
$$

Setting $$\frac{dA}{dx} = 0$$, solve for x:

By symmetry or further solving, maximum occurs when $$x=\frac{\sqrt{200}}{2}=5\sqrt{2}$$. Therefore,

Maximum area $$A_{max}$$:
$$
A_{max}= x\sqrt{100 – x^2} = 50
$$

Key Concept: Basic Interval Identification

c) $$(2, ∞)$$
[Solution Description]
To find intervals of increase or decrease, we first find the derivative $$f'(x)$$.

The derivative $$f'(x) = \frac{d}{dx}(2x^2 – 8x + 5) = 4x – 8$$.

Set $$f'(x) > 0$$ for increasing:
\begin{align*}
4x – 8 &> 0 \\
4x &> 8 \\
x &> 2
\end{align*}
Thus, the function is increasing in the interval $$(2, ∞)$$.

Key Concept: Constant Function

c) Constant
[Solution Description] If $$f'(x) = 0$$ for all x in (a, b), then the function is constant on that interval.

Key Concept: Non-differentiable Points

b) $$x = -1$$

[Solution Description]

A function involving absolute values like $$h(x) = |x^2 – 1|$$ can be non-differentiable at points where the expression inside the absolute value is zero.

Solve $$x^2 – 1 = 0$$:

$$x^2 = 1 \Rightarrow x = \pm 1.$$

Check which values lie within the interval $$[-2, 3]$$:

$$x = -1 \quad \text{and} \quad x = 1.$$

Therefore, the function is not differentiable at $$x = -1$$ and $$x = 1$$. However, if asked for a single point, either could be correct based on context.