Answer :
y = sin x + cos x
the tangent will be horizontal when the derivative is 0
y ' = cos x - sin x
0 = cos x - sin x
sin x = cos x
tan x = 1
this happens when x = pi/4 or 5pi/4
y(pi/4) = sqrt(2) / 2 + sqrt(2) / 2 = sqrt(2)
y(5pi/4) = -sqrt(2) / 2 + -sqrt(2) / 2 = -sqrt(2)
the two points are: (pi/4 , srqt(2)) and (5pi/4 , -srqt(2))
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the tangent will be horizontal when the derivative is 0
y ' = cos x - sin x
0 = cos x - sin x
sin x = cos x
tan x = 1
this happens when x = pi/4 or 5pi/4
y(pi/4) = sqrt(2) / 2 + sqrt(2) / 2 = sqrt(2)
y(5pi/4) = -sqrt(2) / 2 + -sqrt(2) / 2 = -sqrt(2)
the two points are: (pi/4 , srqt(2)) and (5pi/4 , -srqt(2))
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
The required values are,
[tex]x=\frac{\pi}{4},\frac{3\pi}{4} ,\frac{5\pi}{4}[/tex]
Given function is,
[tex]y=sinx cosx[/tex]
Tangent to the graph:
The tangent to the graph [tex]y=f(x)[/tex] is horizontal at [tex]x=a[/tex] if [tex]\frac{dy}{dx} =0[/tex] at [tex]x=a[/tex]
We can write the given equation as,
[tex]y=sinxcos\\y=\frac{1}{2}(sinxcosx)\\ y=\frac{1}{2}sin2x[/tex]
Now, differentiating the given function with respect to [tex]x[/tex].
[tex]\frac{dy}{dx} =\frac{1}{2} (2cos2x)\\\frac{dy}{dx} =cos2x\\\frac{dy}{dx} =0\\cos2x=0\\x=\frac{\pi}{4} ,\frac{3\pi}{4} ,\frac{5\pi}{4} ,\frac{7\pi}{4} ,\frac{9\pi}{4} \\y=\frac{\pi}{4} ,\frac{3\pi}{4} ,\frac{5\pi}{4}[/tex]
Learn more about a tangent to the graph:
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