What is the tangent of f(x)=x^2-1,\at x=3 ?

The tangent of f(x)=x^2-1,\at x=3 is y=6x-10

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tangent of f(x)=x^2-1,\at x=3

tangent of

f (x) = x^{2} - 1, at x = 3

y = 6 x - 10

Solution steps

Find the tangent point:

(3, 8)

Find the slope of

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

Compute the slope of

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

x = 3 : m = 6

Find the line with slope m=

6

and passing through

(3, 8) : y = 6 x - 10

y = 6 x - 10

x \to \infty lim (\frac{1}{3})

\frac{d y}{d x} + 2 y - 8 e^{- x} = 0, y (0) = 7

x y^{^{'}} + (1 + x cot (x)) y = 0

\int_{0}^{1} (e^{x}) - (x e^{x^{2}}) d x

x \to \infty lim (2 g (x) x)

What is the tangent of f(x)=x^2-1,\at x=3 ?
The tangent of f(x)=x^2-1,\at x=3 is y=6x-10