let f be the function defined by f(x)=3X^5 -5X^3 +2
a) on what interval is f increasing? b) on what interval is the graph of f concave upward?
c)Write the equation of each horizontal line tangent to the graph of f
So is the answer y=0, y=2, and y=4?
Oops. f'' = 30x(2x^2 - 1)
so -1/√2 < x < 0 or x > 1/√2
a) The function is increasing on the interval where the derivative is positive. Let's find the derivative of f(x):
f'(x) = d/dx (3x^5 - 5x^3 + 2)
= 15x^4 - 15x^2
To find where f'(x) > 0, we need to solve the inequality:
15x^4 - 15x^2 > 0
Factoring out the common factor of 15x^2, we get:
15x^2(x^2 - 1) > 0
Now, we have two factors: 15x^2 > 0 and x^2 - 1 > 0.
For 15x^2 > 0, the quadratic is always positive for x ≠ 0.
For x^2 - 1 > 0, we solve the inequality:
x^2 - 1 > 0
(x - 1)(x + 1) > 0
This means x > 1 or x < -1.
Therefore, the function is increasing on the intervals: (-∞, -1) and (1, +∞).
b) The graph of f is concave upward when the second derivative is positive. Let's find the second derivative of f(x):
f''(x) = d^2/dx^2 (15x^4 - 15x^2)
= 60x^3 - 30x
To find where f''(x) > 0, we need to solve the inequality:
60x^3 - 30x > 0
Factoring out the common factor of 30x, we get:
30x(x^2 - 1) > 0
Now, we have two factors: 30x > 0 and x^2 - 1 > 0.
For 30x > 0, the quadratic is always positive for x ≠ 0.
For x^2 - 1 > 0, we have x < -1 or -1 < x < 1 or x > 1.
Therefore, the graph of f is concave upward on the intervals: (-∞, -1) and (0, 1).
c) To find the equations of horizontal lines tangent to the graph of f, we need to find the x-values where f'(x) = 0.
15x^4 - 15x^2 = 0
Factoring out the common factor of 15x^2, we get:
15x^2(x^2 - 1) = 0
Setting each factor equal to zero, we have:
15x^2 = 0 --> x = 0
x^2 - 1 = 0 --> x = ±1
Therefore, the horizontal lines tangent to the graph of f have equations:
y = f(0) = 2 at x = 0
y = f(1) = 0 at x = 1
y = f(-1) = 8 at x = -1
To determine the intervals of increase and concavity, we need to analyze the properties of the function f(x)=3x^5 - 5x^3 + 2 and its derivatives.
a) Increasing intervals:
To identify where the function is increasing, we need to find the intervals where the derivative of f(x) is positive.
Step 1: Find the derivative of f(x) with respect to x:
f'(x) = 15x^4 - 15x^2
Step 2: Set the derivative f'(x) greater than zero and solve for x:
15x^4 - 15x^2 > 0
Step 3: Factor the equation:
15x^2(x^2 - 1) > 0
Step 4: Determine when each factor is greater than zero:
For 15x^2 > 0, x can be any real number since the square of a real number is always positive.
For x^2 - 1 > 0, we solve x^2 - 1 = 0, which gives x = -1 and x = 1. We now have three intervals: (-∞, -1), (-1, 1), and (1, ∞).
Thus, f(x) is increasing on the intervals (-∞, -1) and (1, ∞).
b) Concave upward intervals:
To determine where the graph of f(x) concaves upward, we need to find where the second derivative of f(x) is positive.
Step 1: Find the second derivative of f(x):
f''(x) = 60x^3 - 30x
Step 2: Set the second derivative f''(x) greater than zero and solve for x:
60x^3 - 30x > 0
Step 3: Factor the equation:
30x(x^2 - 1) > 0
Step 4: Determine when each factor is greater than zero:
For 30x > 0, we find x > 0 since dividing a positive number by a positive number gives a positive result.
For x^2 - 1 > 0, we already know the solutions from part a: (-∞, -1), (-1, 1), and (1, ∞).
Thus, f(x) is concave upward on the interval (0, ∞).
c) Horizontal lines tangent to the graph of f(x):
To find horizontal lines tangent to the graph of f(x), we need to determine the horizontal line equations where they have the same slope as the function.
Step 1: Find the slope of f(x):
The slope of f(x) is given by its derivative f'(x) found in part a.
f'(x) = 15x^4 - 15x^2
Step 2: Set the slope f'(x) equal to zero and solve for x:
15x^4 - 15x^2 = 0
Step 3: Factor the equation:
15x^2(x^2 - 1) = 0
Step 4: Determine when each factor is equal to zero:
For 15x^2 = 0, x = 0 since the square of zero is zero.
For x^2 - 1 = 0, we already know the solutions from part a: x = -1 and x = 1.
The horizontal line equations tangent to the graph of f(x) are:
y = f(0)
y = f(-1)
y = f(1)
Substitute these x-values into the function f(x)=3x^5 - 5x^3 + 2 to obtain the corresponding y-values.
a) That would be where the derivative
f'(x) = 15x^4 -15x^2 > 0
x^2*(x^2-1) >0
Since x^2 must be positive or zero,
(x+1)(x-1) > 0
x > 1 or x<-1
b) That would be where f"(x) > 0
c) Horizontal tangents would be where f'(x) = 0.
Find those x and y values.
f is increasing when f' is positive
f' = 15x^4 - 15x^2 = 15x^2 (x^2-1)
So, f' > 0 when |x| > 1
f is concave upward when f'' is positive
f'' = 60x^3 - 30x = 30x(2x-1)
So, f'' > 0 when x < 0 or x > 1/2
Horizontal lines have slope=0. So, we want places where f'(x) = 0
15x^2 (x^2 - 1) = 0
x = -1, 0, 1
The horizontal lines are
y=f(-1)
y=f(0)
y=f(1)
evaluate f(x) at those points to get your lines.