Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using interval notation. If the answer cannot be expressed as an interval, enter EMPTY or ∅.)
f(x) = 5x2 + 9x + 8
This is just a parabola
When a function increases, its derivative is positive.
When a function decreases , that derivative is negative
if y = 5x^2 + 9x + 8
dy/dx = 10x + 9
When 10x+9 > 0
x > -9/10
when 10x + 9 < 0
x < -9/10
You do the interval notation stuff, I am "old school" and I use
the above notation.
( bet you the x of the vertex is -9/10)
To find the intervals where the function is increasing or decreasing, we need to analyze the sign of the derivative of the function.
First, we need to find the derivative of f(x):
f'(x) = d/dx (5x^2 + 9x + 8)
= 10x + 9
Now, we can determine the intervals where the function is increasing or decreasing by looking at the sign of the derivative.
To find where the function is increasing, we need to find where the derivative is positive. So we set f'(x) > 0 and solve for x:
10x + 9 > 0
10x > -9
x > -9/10
Therefore, the function is increasing for x > -9/10. The interval where the function is increasing is (-(9/10), ∞).
To find where the function is decreasing, we need to find where the derivative is negative. So we set f'(x) < 0 and solve for x:
10x + 9 < 0
10x < -9
x < -9/10
Therefore, the function is decreasing for x < -9/10. The interval where the function is decreasing is (-∞, -(9/10)).
In summary:
The function f(x) = 5x^2 + 9x + 8 is increasing for x > -9/10 and decreasing for x < -9/10.
To determine the intervals of increase and decrease for the given function, we need to analyze its derivative.
First, we find the derivative of the function f(x) = 5x^2 + 9x + 8.
The derivative, denoted as f'(x), represents the rate of change or slope of the original function.
Using the power rule of differentiation, we find:
f'(x) = 10x + 9
Now, to determine the intervals where the function is increasing or decreasing, we need to examine the sign of the derivative.
1. For the function to be increasing, its derivative f'(x) should be positive.
2. For the function to be decreasing, its derivative f'(x) should be negative.
To find the intervals where the derivative is positive or negative, we set f'(x) = 0 and solve for x:
10x + 9 = 0
10x = -9
x = -9/10
Now, we have a critical point at x = -9/10. We can use the critical point to test intervals on a number line.
Considering x < -9/10:
- Plugging in a value smaller than -9/10 into the derivative f'(x) = 10x + 9:
- For example, x = -1
f'(-1) = 10(-1) + 9 = -1
- Since the derivative is negative, the original function is decreasing on this interval.
Considering x > -9/10:
- Plugging in a value larger than -9/10 into the derivative f'(x) = 10x + 9:
- For example, x = 0
f'(0) = 10(0) + 9 = 9
- Since the derivative is positive, the original function is increasing on this interval.
Therefore, the intervals of increase and decrease for the function f(x) = 5x^2 + 9x + 8 are:
Increasing interval: (-∞, -9/10) ∪ (empty or ∅)
Decreasing interval: (-9/10, +∞)