# If k is greater than or equal to 1, the graphs of y=sinx and y=k(e^(-x)) intersect for x is greater than or equal to 0.A) Find the smallest value of k for which the graphs are tangent. k= B) What...

If k is greater than or equal to 1, the graphs of y=sinx and y=k(e^(-x)) intersect for x is greater than or equal to 0.

A) Find the smallest value of k for which the graphs are tangent.

k=

B) What are the coordinates of the point of tangency?

x=

y=

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### 1 Answer

Given the functions `f(x)=ke^(-x),k>1` and `g(x)=sinx` we want to find the smallest `k` such that the graphs are tangent.

(1) For the graphs to intersect you need `f(x)=g(x)` :

`ke^(-x)=sinx`

(2) For the graphs to be tangent, they must share a tangent line. Thus the first derivative, which gives the slope of the tangent line, must be the same at some point or `f'(c)=g'(c)` .

(3) The derivatives:

`f'(x)=-ke^(-x)`

`g'(x)=cosx`

(4) Now we have a set of simultaneous equations. At the point of intersection `f(x)=g(x)` , and if the graphs are tangent at that point then `f'(x)=g'(x)` . So:

`ke^(-x)=sinx`

`-ke^(-x)=cosx`

`=>sinx=-cosx`

`=>x=(3pi)/4+npi,n in ZZ`

(5) **Let `x=(3pi)/4` . The slope of the tangent line for both curves is `m=(-sqrt(2))/2` and the shared point is `((3pi)/4,(sqrt(2))/2)` **

**At this point `k=sqrt(2)/2e^((3pi)/4)` **

The graphs:

The straight line shown is the tangent line:

`y=(-sqrt(2))/2(x-(3pi)/4)+sqrt(2)/2`