By applying differentiation,

Question

sketch the graph of
k(x) = 20x
3 − 3x
5
.

Answer 1

To sketch the graph of k(x) = 20x^3 - 3x^5, we first need to find the critical points and the behavior of the function at those points.

Taking the derivative of k(x) with respect to x, we get:

k'(x) = 60x^2 - 15x^4

Setting k'(x) = 0, we get:

60x^2 - 15x^4 = 0

15x^2(4 - x^2) = 0

x = 0 or x = ±2

These are the critical points of k(x). To determine the behavior of the function at these points, we can use the first derivative test.

For x < -2 or x > 2, k'(x) is positive, indicating that k(x) is increasing.

For -2 < x < 0, k'(x) is negative, indicating that k(x) is decreasing.

For 0 < x < 2, k'(x) is positive, indicating that k(x) is increasing.

We can also find the inflection points of k(x) by taking the second derivative:

k''(x) = 120x - 60x^3

Setting k''(x) = 0, we get:

120x - 60x^3 = 0

x(2 - x)(2 + x) = 0

x = 0 or x = ±2

These are the inflection points of k(x). To determine the concavity of the function at these points, we can use the second derivative test.

For x < -2 or x > 2, k''(x) is positive, indicating that k(x) is concave up.

For -2 < x < 0, k''(x) is negative, indicating that k(x) is concave down.

For 0 < x < 2, k''(x) is positive, indicating that k(x) is concave up.

Finally, we can sketch the graph of k(x) using this information:

- At x = 0, k(x) has a local maximum.
- At x = ±2, k(x) has inflection points.
- The function is increasing from negative infinity to x = -2, decreasing from x = -2 to x = 0, increasing from x = 0 to x = 2, and then increasing from x = 2 to positive infinity.
- The function is concave down from negative infinity to x = -2, concave up from x = -2 to x = 0, concave down from x = 0 to x = 2, and concave up from x = 2 to positive infinity.

Based on this information, we can sketch the graph of k(x) as follows:


Answer 2

Doctors have to attend to a
patient on top of a building
using a 50 m ladder before
s/he is evacuated to a hospital.
The ladder is placed against
the building. The top of the
ladder is sliding down the
building at the rate of 2
m/min. Determine the rate at
which the base of the ladder is
moving away from the building
at the instant that the base is
30 m from the building.
Differentiation

Answer 3

We can use related rates to solve this problem. Let's call the distance from the base of the ladder to the building "x" and the distance from the top of the ladder to the ground "y". We want to find the rate at which x is changing when x = 30 m.

We know that the ladder is 50 m long and is sliding down the building at a rate of 2 m/min. That means that dy/dt = -2 m/min (we have a negative sign because y is decreasing).

We want to find dx/dt when x = 30 m. To do this, we need to relate x and y using the Pythagorean theorem:

x^2 + y^2 = 50^2

Differentiating both sides with respect to time (t) using the chain rule, we get:

2x(dx/dt) + 2y(dy/dt) = 0

Plugging in y = sqrt(50^2 - x^2) and dy/dt = -2, we get:

2x(dx/dt) + 2sqrt(50^2 - x^2)(-2) = 0

Simplifying and solving for dx/dt, we get:

dx/dt = (sqrt(50^2 - x^2))(2/x)

Now we can plug in x = 30 to find the rate at which x is changing when the base of the ladder is 30 m from the building:

dx/dt = (sqrt(50^2 - 30^2))(2/30)

dx/dt ≈ 1.75 m/min

Therefore, the base of the ladder is moving away from the building at a rate of approximately 1.75 m/min when it is 30 m from the building.