We are given three problems: a) Sketch the graph of the function $f(x) = x^3 - 12x^2 + 45x - 40$. b) Determine the area enclosed by the line $y = 2x + 3$, the x-axis, and the vertical lines $x = 1$ and $x = 4$. c) Given the velocity of a body as a function of time $v(t) = 2t^2 + 5$ m/s, find the distance it moves in the time interval from $t = 0$ to $t = 4$ seconds.

1. Problem Description

We are given three problems:

a) Sketch the graph of the function $f(x) = x^3 - 12x^2 + 45x - 40$.

b) Determine the area enclosed by the line $y = 2x + 3$, the x-axis, and the vertical lines $x = 1$ and $x = 4$.

c) Given the velocity of a body as a function of time $v(t) = 2t^2 + 5$ m/s, find the distance it moves in the time interval from $t = 0$ to $t = 4$ seconds.

2. Solution Steps

a) To sketch the graph of $f(x) = x^3 - 12x^2 + 45x - 40$, we can find the first derivative to find the critical points:

$f'(x) = 3x^2 - 24x + 45$.

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

$3x^2 - 24x + 45 = 0$

$x^2 - 8x + 15 = 0$

$(x - 3)(x - 5) = 0$

So, the critical points are $x = 3$ and $x = 5$.

The second derivative is $f''(x) = 6x - 24$.

At $x = 3$, $f''(3) = 6(3) - 24 = 18 - 24 = -6 < 0$, so $x = 3$ is a local maximum. $f(3) = (3)^3 - 12(3)^2 + 45(3) - 40 = 27 - 108 + 135 - 40 = 14$. The local maximum is at $(3, 14)$.

At $x = 5$, $f''(5) = 6(5) - 24 = 30 - 24 = 6 > 0$, so $x = 5$ is a local minimum. $f(5) = (5)^3 - 12(5)^2 + 45(5) - 40 = 125 - 300 + 225 - 40 = 10$. The local minimum is at $(5, 10)$.

Also, we can find the y-intercept by setting $x=0$: $f(0) = -40$. The y-intercept is at $(0, -40)$.

The graph is a cubic function with a local maximum at $(3, 14)$ and a local minimum at $(5, 10)$. It crosses the y-axis at $(0, -40)$. It increases for $x<3$, decreases for $3<x<5$, and increases for $x>5$. We also can notice that $f(1) = 1 - 12 + 45 - 40 = -6$ and $f(2) = 8 - 48 + 90 - 40 = 10$.

b) To determine the area enclosed by $y = 2x + 3$, the x-axis, and the lines $x = 1$ and $x = 4$, we need to integrate the function $y = 2x + 3$ from $x = 1$ to $x = 4$.

Area = $\int_{1}^{4} (2x + 3) dx = [x^2 + 3x]_{1}^{4} = (4^2 + 3(4)) - (1^2 + 3(1)) = (16 + 12) - (1 + 3) = 28 - 4 = 24$.

c) To find the distance the body moves from $t = 0$ to $t = 4$, we need to integrate the velocity function $v(t) = 2t^2 + 5$ from $t = 0$ to $t = 4$.

Distance = $\int_{0}^{4} (2t^2 + 5) dt = [\frac{2}{3}t^3 + 5t]_{0}^{4} = (\frac{2}{3}(4)^3 + 5(4)) - (\frac{2}{3}(0)^3 + 5(0)) = (\frac{2}{3}(64) + 20) - 0 = \frac{128}{3} + 20 = \frac{128}{3} + \frac{60}{3} = \frac{188}{3}$.

3. Final Answer

a) Sketch: The graph of $f(x) = x^3 - 12x^2 + 45x - 40$ has a local maximum at $(3, 14)$, a local minimum at $(5, 10)$ and y-intercept at $(0, -40)$.

b) Area =

2

4.

c) Distance = $\frac{188}{3}$ meters.