To determine
To find:
x at which curve y=sinx+cosx 0≤x≤π/2 is horizontal.
Answer
At
π4, 2and (5π4,2)
the tangent to curve y=sinx+cosx 0≤x≤π/2 is horizontal.
Explanation
1) Formula:
i. Sum rule:
ddxfx+gx=ddxfx+ddxg(x)
ii. ddxsinx=cosx
iii. ddxcosx=-sinx
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=sinx+cosx 0≤x≤π/2
3) Calculation:
Differentiate given equation of curve with respect to x,
f'x=ddx[sinx+cosx]
Using sum rule,
ddxsinx+cosx=ddxsinx+ddxcosx=cosx-sinx
Tangent line on curve is horizontal if f'x=0
f'x=0
cosx-sinx=0
cosx=sinx
tanx=1
⇒x=tan-11
On 0≤x≤π/2
x=tan-11⇒x=π4,5π4
When x=π4, y=sinπ4+cosπ4= 2, so the corresponding point is
π4, 2
When x=5π4, y=sinπ4+cosπ4= -2, so the corresponding point is
(5π4, -2)
Conclusion:
At (π4,2) and ( 5π4, -2 )
The curve y=sinx+cosx 0≤x≤π/2 is horizontal.