Can anybody please help me with this derivatives problem about graphs?

Question

Show that the curves y= e^(-x) and y= -e^(-x) touch the curve
y=e^(-x)*sinx at its inflection points.

Answer 1

Inflection points are points where the second derivative is equal to zero. Start by taking the derivative of y to get:
y' = -e^(-x)*sinx + e^(-x)*cosx.

Now take the derivative of this to get:
y'' = e^(-x)*sinx + -e^(-x)*cosx -e^(-x)*cosx - e^(-x)*sinx.
= -2e^(-x)cosx

Set this function equal to zero to find the inflection points.
0 = -2e^(-x)cosx
We see that since e to any power is always positive, the only time for this to be zero is when cosx = 0. Cosine is equal to zero at the points pi/2 and 3pi/2.

Now go back to the original equation, y = e^(-x)*sinx, to find what the y values are at the points pi/2 and 3pi/2.
y = e^(-pi/2)*sin(pi/2) = e^(-pi/2)
y = e^(-3pi/2)*sin(3pi/2) = -e^(-3pi/2)

Now plug these values in for y = e^(-x) and y = -e^(-x) to find that they do infact touch the graph at its inflection points.

Answer 2

e^-x = e^-x sinx
e^-x (1 - sinx ) = 0
Hence sin x = 1 --> x = pi/2 which is an inflection point
Similarly e^-x(1+sinx) = 0
sinx = -1 --> x = 3pi/2 which is also an inflection point

No derivatives needed.

Answer 3

i am not good in derivatives

Answer 4

No, I can't help, sorry.