What is the tangent of f(x)=x^2*e^{x^2-1},\at x=-1 ?

The tangent of f(x)=x^2*e^{x^2-1},\at x=-1 is y=-4x-3

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tangent of f(x)=x^2*e^{x^2-1},\at x=-1

tangent of

f (x) = x^{2} \cdot e^{x^{2} - 1}, at x = - 1

y = - 4 x - 3

Solution steps

Find the tangent point:

(- 1, 1)

Find the slope of

f (x) = x^{2} e^{x^{2} - 1} : \frac{df ( x )}{d x} = 2 e^{x^{2} - 1} x + 2 e^{x^{2} - 1} x^{3}

Compute the slope of

f (x) = x^{2} e^{x^{2} - 1}

x = - 1 : m = - 4

Find the line with slope m=

- 4

and passing through

(- 1, 1) : y = - 4 x - 3

y = - 4 x - 3

\int (x - 1) x^{2} - 2 x d x

\frac{d}{d h} (35 f (h) (3 + 2 h))

(1 + \frac{x}{2})^{- 4}

\int 16 - x d x

3 x^{2} y^{^{''}} + 7 x y^{^{'}} + y = 24 x + 2

What is the tangent of f(x)=x^2*e^{x^2-1},\at x=-1 ?
The tangent of f(x)=x^2*e^{x^2-1},\at x=-1 is y=-4x-3