What is the equation of the line tangent to f(x)=ex2+x at x=−1?

1 Answer
Barry H.
May 12, 2018

Equation of tangent line, y=−x

Explanation:

First we must find the gradient at point x=−1 by differentiating the expression f[x]=ex2+x

By the chain rule ddx[ex2+x]=ex2+xddx[x2+x] =2x+1[ex2+x]. substituting x=−1 into this expression, ddx=[2[−1]+1][e−12−1]=−e0=−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 functionf[x] when x=−1

When x=−1, f[−1] = e[−12]−1 = e0, =1 Thus the equation of the tangent line =[y−1]=−1[x+1], = [y=−x].

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