Which value of x in the interval π2≤x≤π makes the line tangent to the function f(x)=2–√x+2cosx horizontal
To find the value of x that makes the line tangent to the function f(x) = 2 – √x + 2cosx horizontal, we need to find when the derivative of f(x) is equal to zero.
First, let's find the derivative of f(x):
f'(x) = d/dx (2 – √x + 2cosx)
= -1/(2√x) - 2sinx
We set f'(x) = 0 to find the critical points:
-1/(2√x) - 2sinx = 0
Multiplying through by 2√x gives us:
-1 - 2√xsinx = 0
Rearranging, we have:
2√xsinx = -1
Dividing by 2√x gives us:
sinx/√x = -1/2
To find the value of x in the interval π/2 ≤ x ≤ π that satisfies this equation, we can use a graphing calculator or a table of values. However, in this case, we can recognize that sinx/√x is always positive in this interval since sinx is positive and √x is positive. Therefore, there is no solution in the interval π/2 ≤ x ≤ π.
To find the value of x that makes the line tangent to the function f(x) = 2 – √x + 2cos(x) horizontal in the interval π/2 ≤ x ≤ π, we need to find the derivative of the function and set it equal to zero.
Step 1: Find the derivative of f(x) with respect to x.
f'(x) = d/dx [2 – √x + 2cos(x)]
To find the derivative of √x, we can use the power rule:
f'(x) = 0 - (1/2)√x + 2(-sin(x))
Step 2: Set the derivative equal to zero and solve for x.
0 - (1/2)√x + 2(-sin(x)) = 0
-(1/2)√x - 2sin(x) = 0
Step 3: Solve the equation for x.
-(1/2)√x - 2sin(x) = 0
-(1/2)√x = 2sin(x)
√x = -4sin(x)
Square both sides:
x = 16sin^2(x)
We can simplify the equation by using the identity sin^2(x) + cos^2(x) = 1:
x = 16(1 - cos^2(x))
x = 16 - 16cos^2(x)
Step 4: Determine the values of x in the interval π/2 ≤ x ≤ π where the line is tangent to f(x) horizontally.
In this interval, x starts at π/2 and ends at π. We will substitute these values into the equation x = 16 - 16cos^2(x) and check if the line is horizontal.
Substituting x = π/2:
π/2 = 16 - 16cos^2(π/2)
π/2 = 16 - 16(0)
π/2 = 16
The line is not horizontal at x = π/2.
Substituting x = π:
π = 16 - 16cos^2(π)
π = 16 - 16(-1)^2
π = 16 - 16(1)
π = 0
The line is horizontal at x = π.
Therefore, the value of x in the interval π/2 ≤ x ≤ π that makes the line tangent to the function f(x) horizontal is x = π.
To find the value of x in the interval π/2 ≤ x ≤ π that makes the line tangent to the function f(x) = 2 - √x + 2cos(x) horizontal, we need to solve for the derivative of the function and set it equal to zero.
Step 1: Find the derivative of the function f(x).
f(x) = 2 - √x + 2cos(x)
f'(x) = d/dx(2 - √x + 2cos(x))
= -1/2√x - 2sin(x)
Step 2: Set the derivative equal to zero and solve for x.
-1/2√x - 2sin(x) = 0
-1/2√x = 2sin(x)
√x = -4sin(x)
Square both sides:
x = 16sin^2(x)
Step 3: Simplify the equation.
Using the trigonometric identity sin^2(x) + cos^2(x) = 1, we can rewrite the equation as:
x = 16(1 - cos^2(x))
x = 16 - 16cos^2(x)
Step 4: Solve the equation for x in the given interval.
As the equation x = 16 - 16cos^2(x) cannot be solved algebraically, we need to use numerical methods to find the answer. In this case, let's use an iterative method like the Newton-Raphson method.
Starting with an initial guess within the given interval, such as x = π/2, we can use the following iteration formula:
x(n+1) = x(n) - f(x(n))/f'(x(n))
By repeating this process, we can approximate the value of x that satisfies the equation.
Note: Since this is a complex calculation, it is recommended to use appropriate software or tools to solve the equation numerically.