How do you find the points where the graph of the function #
f(x) = 1+sinxcosx# has horizontal tangents and what is the equation?

Question

Luke Alvarez · Accepted Answer

#x = pi/4 + pin# and #(3pi)/4 + pin#

Differentiate.

#f'(x) = cosx(cosx) + sinx(-sinx)#

#f'(x) = cos^2x- sin^2x#

#f'(x) = cos2x#

The tangent will be horizontal when it has a slope of #0#. Set the derivative to #0# and solve:

#0 = cos2x#

#2x = pi/2 and (3pi)/2#

#x = pi/4 and (3pi)/4#

If you add in periodicity, you have:

#x = pi/4 + pin# and #(3pi)/4 + pin#

Because #cos2x# has a period of #pi#.

Hopefully this helps!

Cameron Alexander · Answer

undefined To find the points where the graph of the function f(x) = 1 + sin(x)cos(x) has horizontal tangents, we need to find the values of x where the derivative of the function is equal to zero.

First, we find the derivative of f(x) using the product rule:

f'(x) = cos(x)cos(x) + sin(x)(-sin(x)) = cos^2(x) - sin^2(x)

Next, we set f'(x) equal to zero and solve for x:

cos^2(x) - sin^2(x) = 0

Using the trigonometric identity cos^2(x) - sin^2(x) = cos(2x), we can rewrite the equation as:

cos(2x) = 0

To find the values of x that satisfy this equation, we need to find the solutions for cos(2x) = 0.

The solutions for cos(2x) = 0 occur when 2x is equal to π/2 + nπ, where n is an integer.

Solving for x, we have:

2x = π/2 + nπ

x = (π/2 + nπ)/2

Therefore, the points where the graph of f(x) = 1 + sin(x)cos(x) has horizontal tangents are given by x = (π/2 + nπ)/2, where n is an integer.

How do you find the points where the graph of the function # f(x) = 1+sinxcosx# has horizontal tangents and what is the equation?