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

The tangent of f(x)=x^2-1,\at x=2 is y=4x-5

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

tangent of

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

y = 4 x - 5

Solution steps

Find the tangent point:

(2, 3)

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 = 2 : m = 4

Find the line with slope m=

4

and passing through

(2, 3) : y = 4 x - 5

y = 4 x - 5

y = e^{x}, y = e^{2 x}, x = 0, x = ln (2)

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

f (x) = 4 x^{2} - 4 x + 1, at x = 0

\int 5^{2} - x^{2} d x

\frac{\partial}{\partial x} (x tan (x))

What is the tangent of f(x)=x^2-1,\at x=2 ?
The tangent of f(x)=x^2-1,\at x=2 is y=4x-5