Extrema of Functions

Approach:

Decide the extrema by graphing and comparing values.

Steps:

1. Sketch the graph: This is a quadratic function whose graph is an upward-opening parabola.

2. Identify the extremum point: Since the parabola opens upward, the function has a minimum point and no maximum point.

3. Find the minimum point: For a quadratic $f(x) = ax^2 + bx + c$, the axis of symmetry is $x = -\frac{b}{2a}$ $x = -\frac{-4}{2 \times 1} = 2$

4. Compute the extremum: $f(2) = 2^2 - 4 \times 2 + 3 = 4 - 8 + 3 = -1$

5. Verify: Take $x = 1$, $f(1) = 1 - 4 + 3 = 0 > -1$ Take $x = 3$, $f(3) = 9 - 12 + 3 = 0 > -1$ Confirms $x = 2$ is a minimum point.

Answer: The function attains a local minimum $-1$ at $x = 2$.

Approach:

Decide the extrema by graphing and comparing values.

Steps:

1. Sketch the graph: This is a quadratic function whose graph is a downward-opening parabola.

2. Identify the extremum point: Since the parabola opens downward, the function has a maximum point and no minimum point.

3. Find the maximum point: For a quadratic $f(x) = ax^2 + bx + c$, the axis of symmetry is $x = -\frac{b}{2a}$ $x = -\frac{2}{2 \times (-1)} = 1$

4. Compute the extremum: $f(1) = -1^2 + 2 \times 1 + 1 = -1 + 2 + 1 = 2$

5. Verify: Take $x = 0$, $f(0) = 0 + 0 + 1 = 1 < 2$ Take $x = 2$, $f(2) = -4 + 4 + 1 = 1 < 2$ Confirms $x = 1$ is a maximum point.

Answer: The function attains a local maximum $2$ at $x = 1$.

Approach:

Build a function relationship and find the extremum by graphing.

Steps:

1. Build the function: Let the length be $x$ and the width be $y$. Perimeter: $2(x + y) = 20$, so $y = 10 - x$. Area: $S = xy = x(10 - x) = 10x - x^2$.

2. Analyze the function: This is a quadratic $S(x) = -x^2 + 10x$, opening downward, so it has a maximum.

3. Find the maximum point: Axis of symmetry: $x = -\frac{10}{2 \times (-1)} = 5$.

4. Compute the maximum: $S(5) = 10 \times 5 - 5^2 = 50 - 25 = 25$.

5. Verify: When $x = 4$, $S(4) = 40 - 16 = 24 < 25$. When $x = 6$, $S(6) = 60 - 36 = 24 < 25$.

Answer: When the rectangle is a square with sides $5$ cm, the area reaches its maximum of $25$ cm².

Approach:

Decide the extrema by graphing and comparing values.

Steps:

1. Sketch the graph: This is a quadratic function whose graph is an upward-opening parabola.

2. Identify the extremum point: Since the parabola opens upward, the function has a minimum point and no maximum point.

3. Find the minimum point: Axis of symmetry: $x = -\frac{b}{2a}$, so $x = -\frac{2}{2 \times 1} = -1$.

4. Compute the extremum: $f(-1) = (-1)^2 + 2 \times (-1) + 1 = 1 - 2 + 1 = 0$.

5. Verify: Take $x = -2$, $f(-2) = 4 - 4 + 1 = 1 > 0$. Take $x = 0$, $f(0) = 0 + 0 + 1 = 1 > 0$. Confirms $x = -1$ is a minimum point.

Answer: The function attains a local minimum $0$ at $x = -1$.

Approach:

Decide the extrema by graphing and comparing values.

Steps:

1. Sketch the graph: This is a quadratic function whose graph is a downward-opening parabola.

2. Identify the extremum point: Since the parabola opens downward, the function has a maximum point and no minimum point.

3. Find the maximum point: Axis of symmetry: $x = -\frac{6}{2 \times (-1)} = 3$.

4. Compute the extremum: $f(3) = -3^2 + 6 \times 3 - 5 = -9 + 18 - 5 = 4$.

5. Verify: Take $x = 2$, $f(2) = -4 + 12 - 5 = 3 < 4$. Take $x = 4$, $f(4) = -16 + 24 - 5 = 3 < 4$. Confirms $x = 3$ is a maximum point.

Answer: The function attains a local maximum $4$ at $x = 3$.

Approach:

This question tests extremum determination for an integral-defined function. Differentiate $f(x)$, check whether $x=0$ is an extremum point, and compute the extremum.

Steps:

1. By the fundamental theorem of calculus, $f'(x) = e^{x^2} \sin x$.
2. Compute $f'(0)$: $f'(0) = e^{0} \sin 0 = 1 \times 0 = 0$, so the derivative vanishes at $x=0$, which is a possible extremum point.
3. Differentiate again: $$f''(x) = \frac{d}{dx}[e^{x^2} \sin x] = 2x e^{x^2} \sin x + e^{x^2} \cos x$$
4. Compute $f''(0)$: $$f''(0) = 2 \times 0 \times 1 \times 0 + 1 \times 1 = 0 + 1 = 1 > 0$$ so $x=0$ is a local minimum point.
5. Compute the minimum: $$f(0) = \int_0^0 e^{t^2} \sin t\,dt = 0$$

Answer: $x=0$ is a local minimum point of $f(x)$, with local minimum value $0$.