The values of 'a' for which the functions y=a^x and y=1+x are tangent at x=0 is definitively a=e. This conclusion is reached by setting the derivatives of both functions equal at the point of tangency, resulting in the equation a^xln(a)=1. Substituting x=0 leads to ln(a)=1, confirming that a=e is the correct solution. Additionally, it is essential to verify that both functions yield the same value at x=0, which is a crucial step in confirming tangency.
PREREQUISITESStudents studying calculus, particularly those focusing on derivatives and tangents, as well as educators looking for examples of function behavior at specific points.
Painguy said:Homework Statement
For what values of a are y=a^x and y=1+x tangent at x=0? ExplainHomework Equations
y1=1+x
y2=a^x
y2'=a^xln(a)
y1'=1The Attempt at a Solution
Since both equations are tangent at x=0 i set their derivatives equal to each other in hopes of getting a a^xln(a)=1 I then substitute x in that equation with 0 and end up with ln(a)=1.
I end up with a = e. Is this correct?
No i did not, but in hindsight i should have done that. Thanks for the reminder :)Dick said:It sure is correct. You also checked that y1(0)=y2(0), yes?