By applying differentiation,
sketch the graph of
k(x) = 20x
3 − 3x
5
.
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:
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
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.
To sketch the graph of k(x) = 20x^3 - 3x^5 using differentiation, we need to find the critical points, determine the intervals of increasing and decreasing, and analyze the concavity of the function.
Step 1: Find the derivative of k(x)
The derivative of k(x) is found by applying the power rule of differentiation:
k'(x) = 60x^2 - 15x^4
Step 2: Find the critical points
The critical points occur when the derivative equals zero or is undefined. To find these points, set k'(x) = 0 and solve for x:
60x^2 - 15x^4 = 0
Factor out common terms to get:
x^2 (60 - 15x^2) = 0
From here, we have two possible solutions: x = 0 or x = ±√(4).
Step 3: Determine intervals of increasing and decreasing
To determine the intervals of increasing and decreasing, we need to analyze the sign of the derivative.
For x < -√(4), k'(x) is positive, indicating that the function is increasing.
Between -√(4) and 0, k'(x) is negative, indicating that the function is decreasing.
Between 0 and √(4), k'(x) is positive, indicating that the function is increasing.
For x > √(4), k'(x) is negative, indicating that the function is decreasing.
Step 4: Analyze the concavity
To analyze the concavity, we need to find the second derivative of k(x):
k''(x) = 120x - 60x^3
Step 5: Determine the inflection points
The inflection points occur when the second derivative equals zero or is undefined. To find these points, set k''(x) = 0 and solve for x:
120x - 60x^3 = 0
Factor out common terms to get:
x (120 - 60x^2) = 0
From here, we have two possible solutions: x = 0 or x = ±√(2).
Step 6: Sketch the graph
Based on the information obtained, we can now sketch the graph of k(x) = 20x^3 - 3x^5:
- The graph passes through the origin (0,0)
- The critical points occur at x = 0 and x = ±√(4)
- The inflection points occur at x = 0 and x = ±√(2)
Using this information, plot the points and connect them with smooth curves to create the graph of k(x). Be mindful of the increasing/decreasing intervals and concavity to accurately sketch the graph.