For #f(x)=2^x # what is the equation of the tangent line at #x=-5#?

1 Answer
Oct 29, 2015

#y=ln2/32x+0,14#

Explanation:

The gradient of the tangent to the curve at -5 may be given by the derivative at the point -5.

#therefore d/dx2^x=2^x*ln2#

#therefore d/dx|_(x=-5)=2^(-5)*ln2=ln2/32#

But a tangent is a straight line so has form #y=mx+c#

Since #f(-5)=2^(-5)=1/32#, we may substitute the point #(-5,1/32)# into the equation of the tangent to get :

#1/32=(ln2/32)(-5)+c# from which #c=0,14.

Thus the equation of the required tangent to the curve is
#y=ln2/32x+0,14#