Question

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

Answer 1

`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`