Question

What is the equation of the line tangent to `f(x)=e^(x^2+x)` at `x=-1`?

Answer 1

Equation of tangent line, `y=-x`

Explanation:

First we must find the gradient at point `x=-1` by differentiating the expression `f[x]=e^[x^2+x]`

By the chain rule `d/dx [e^[x^2+x]]= e^[x^2+x]d/dx[x^2+x]` =`2x+1[e^[x^2+x]]`. substituting `x = -1` into this expression, `d/dx=[2[-1]+1][e^[-1^2-1]]`=`-e^0=-1`.

So the gradient is `[-1]`, From the equation of a straight line,

`[y-y1]=-1[x-x1]` [since the gradient is `-1`.] We now need to find the value of `y` from the original function`f[x]` when `x=-1`

When `x=-1`, `f[-1]` = `e^[[-1^2]-1]` = `e^0`, =`1` Thus the equation of the tangent line =`[y-1]=-1[x+1]`, = `[y=-x]`.