Find the coordinates of the stationary points on the curve y= x+4/x determine the nature of the turning points.Find also the set of values of x for which y increases as as x increases.
dy/dx = 1 - 4/x^2 = 0 for turning points
Solve for x, then find the matching y to find the turning points.
Decide if those points are max or mins using the 2nd derivative test.
y will increase if dy/dx is positive
y will decrease if dy/dx is negative
To find the coordinates of the stationary points on the curve \(y = \frac{x+4}{x}\), we need to first calculate the derivative of the function with respect to \(x\) and set it equal to zero to find the critical points.
Step 1: Calculate the derivative of \(y\) with respect to \(x\).
\(y = \frac{x+4}{x}\)
Using the quotient rule, the derivative of \(y\) with respect to \(x\) is:
\(y' = \frac{(1 \cdot x) - (x+4) \cdot 1}{x^2}\)
Simplifying this expression, we get:
\(y' = \frac{-4}{x^2}\)
Step 2: Set the derivative equal to zero to find the critical points.
\(y' = \frac{-4}{x^2} = 0\)
To find the critical points, we need to solve this equation for \(x\).
If \(\frac{-4}{x^2} = 0\), it means that the numerator must be zero:
\(-4 = 0\)
However, \(-4\) is not equal to zero, so there are no critical points for this function.
Step 3: Determine the nature of the turning points.
Since there are no critical points, there are no turning points on the curve.
Step 4: Find the set of values of \(x\) for which \(y\) increases as \(x\) increases.
To find where \(y\) increases as \(x\) increases, we need to examine the sign of the derivative for different ranges of \(x\).
Since the derivative is negative (\(\frac{-4}{x^2} < 0\)) for all values of \(x\), \(y\) is always decreasing as \(x\) increases. Therefore, there is no set of values of \(x\) for which \(y\) increases as \(x\) increases.
To find the coordinates of the stationary points on the curve y = x + 4/x, we need to find the values of x where the derivative of y with respect to x equals zero.
Let's start by finding the derivative of y with respect to x. The function y = x + 4/x can be rewritten as y = x + 4x^(-1). Using the power rule for derivatives, we can find the derivative as:
dy/dx = 1 - 4x^(-2)
Now, we set the derivative equal to zero and solve for x:
1 - 4x^(-2) = 0
To solve this equation, let's multiply both sides by x^2:
x^2 - 4 = 0
Now, we can solve for x by taking the square root of both sides:
x = ±2
So, there are two possible values for x: x = 2 and x = -2.
To determine the nature of the turning points, we need to examine the second derivative. Taking the derivative of the derivative, we get:
d^2y/dx^2 = 8x^(-3)
Now let's evaluate the second derivative at each of the stationary points we found:
For x = 2:
d^2y/dx^2 = 8(2^(-3)) = 1/2
For x = -2:
d^2y/dx^2 = 8(-2^(-3)) = -1/2
So, at x = 2, the second derivative is positive, indicating a minimum turning point, and at x = -2, the second derivative is negative, indicating a maximum turning point.
Next, let's find the set of values of x for which y increases as x increases. In other words, we need to find where the first derivative, dy/dx, is positive.
From our earlier calculation, the derivative is given by:
dy/dx = 1 - 4x^(-2)
To find where dy/dx > 0, we solve the inequality:
1 - 4x^(-2) > 0
Adding 4x^(-2) to both sides, we get:
1 > 4x^(-2)
Taking the reciprocal of both sides, we have:
x^2 > 1/4
Taking the square root of both sides, we get:
x > 1/2 or x < -1/2
Therefore, the set of values of x for which y increases as x increases is x > 1/2 or x < -1/2.
In summary:
Coordinates of the stationary points:
- (2, 6)
- (-2, -2)
Nature of the turning points:
- (2, 6): Minimum turning point
- (-2, -2): Maximum turning point
Set of values of x for which y increases as x increases:
- x > 1/2 or x < -1/2