My question is how do I find inflection points for trig functions?? My original math problem was h(x)=cosx+sinx on an interval of [0, 2 (/pi)] and I dont know how to find the second derivative for a function that has no whole numbers for me to work with. Thank you
#AP Calc hw second derivative
neat jackal
worldly quarry
My question is how do I find inflection points for trig functions?? My original ...
y = sin x
y' = cos x
y'' = - sin x
y''' = -cos x
y'''' = sin x
neat jackal
sorry there was a typo when i typed the function, its h(x)=sinx+cosx. Also can you explain your response? I dont quite understand
leaden current
N.B. $h(x) = \cos{x} + \sin{x} = \sin{x} + \cos{x}$ in the domain $[0, 2\pi]$
Inflection points occur when $$h(x) = 0$$
$$\Rightarrow 0 = \cos{x} + \sin{x}$$
$$\Rightarrow -\cos{x} = \sin{x}$$
$$\Rightarrow -\sin{(x + \frac{\pi}{2})} = \sin{x}$$
Think about the unit circle and how we could get these two to cancel each other out. I suggest drawing them.
This problem may have two solutions, since these functions are cyclic.
crimson gulchBOT
Pigeon
leaden current
The second derivative would produce $-h(x)$.
$$h(x) = \sin{x} + \cos{x}$$
$$\frac{dh(x)}{dx} = \cos{x} + -\sin{x}$$
$$\frac{d^2h(x)}{dx^2} = -\sin{x} - \cos{x}$$