y=5/x-3
find the zeros,relative min and max.the max and min
find the intervals on which yis positive or negative. on which y is decreasing.intervals y concave up or down.points of inflections
I bet you mean
y = 5/(x-3) ????
if so then
approches zero at - infinity and + infinity
undefined at x = 3
negative between x = -oo and x = 3
positive between x = 3 and x = +oo
decreasing for all x except exactly as x = 3
In other words goes down from 0 to -oo as x goes from -oo to 3
then goes down from +oo to 0 as x goes from 3 to +oo
To find the zeros of the given equation y = 5/x - 3, we need to set y to zero and solve for x:
0 = 5/x - 3
To simplify, we multiply both sides of the equation by x:
0 = 5 - 3x
Next, we isolate x by subtracting 5 from both sides of the equation:
3x = 5
Finally, divide both sides by 3 to solve for x:
x = 5/3
So, the zero of the equation is x = 5/3.
To find the relative minimum and maximum, we need to take the derivative of the equation with respect to x and set it equal to zero.
y = 5/x - 3
Taking the derivative:
dy/dx = -5/x^2
Setting dy/dx = 0:
-5/x^2 = 0
Since the denominator cannot be zero, there are no critical points. Therefore, there are no relative minimum or maximum points in this equation.
To find the maximum and minimum values, we can analyze the behavior of the equation. As x approaches positive infinity, y approaches 0. As x approaches negative infinity, y also approaches 0. So the maximum value of y is 0, and the minimum value of y is also 0.
To determine the intervals on which y is positive or negative, we can analyze the behavior of the equation.
When y is positive:
- If x > 5/3, then y > 0.
- If x < 5/3, then y < 0.
When y is negative:
- If x > 5/3, then y < 0.
- If x < 5/3, then y > 0.
To determine the intervals in which y is decreasing, we can analyze the derivative of the equation.
The derivative, dy/dx = -5/x^2, is always negative since the numerator is negative. Therefore, y is decreasing for all values of x.
To determine the intervals where y is concave up or down, we need to analyze the second derivative of the equation, which is the derivative of the derivative.
Taking the second derivative of y:
d^2y/dx^2 = 10/x^3
Since the second derivative is positive for all values of x, except for x = 0, y is concave up for all values of x, except at x = 0.
There are no points of inflection in this equation since the second derivative does not change sign.
In summary:
- Zeros: x = 5/3
- Relative min and max: None
- Maximum value: 0
- Minimum value: 0
- Intervals where y is positive: x < 5/3 and x > 5/3
- Intervals where y is negative: 5/3 < x < 5/3
- Intervals where y is decreasing: All values of x
- Intervals where y is concave up: All values of x, except x = 0
- Points of inflection: None