To determine

To find:

x at which curve y=sin⁡x+cos⁡x   0≤x≤π/2 is horizontal.

Answer

At

π4, 2and (5π4,2) 

the tangent to curve y=sin⁡x+cos⁡x   0≤x≤π/2 is horizontal.

Explanation

1) Formula:

i. Sum rule:

ddxfx+gx=ddxfx+ddxg(x)

ii. ddxsin⁡x=cos⁡x

iii. ddxcos⁡x=-sin⁡x

iv. Equation of tangent line to the curve y=f(x) at point (a,f(a)) is, (y-f(a))=f’(a)(x-a)

2) Given:

y=sin⁡x+cos⁡x   0≤x≤π/2

3) Calculation:

Differentiate given equation of curve with respect to x,

f'x=ddx[sin⁡x+cos⁡x]

Using sum rule,

ddxsin⁡x+cos⁡x=ddxsin⁡x+ddxcos⁡x=cos⁡x-sin⁡x

Tangent line on curve is horizontal if f'x=0

f'x=0

cos⁡x-sin⁡x=0

cos⁡x=sin⁡x

tan⁡x=1

⇒x=tan-1⁡1

On    0≤x≤π/2

x=tan-1⁡1⇒x=π4,5π4

When x=π4, y=sin⁡π4+cos⁡π4= 2, so the corresponding point is

π4, 2

When x=5π4, y=sin⁡π4+cos⁡π4= -2, so the corresponding point is

(5π4, -2) 

Conclusion:

At (π4,2) and ( 5π4, -2 )

The curve y=sin⁡x+cos⁡x   0≤x≤π/2 is horizontal.