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

The tangent of f(x)=x^2-1,\at x=4 is y=8x-17

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tangent of f(x)=x^2-1,\at x=4

tangent of

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

y = 8 x - 17

Solution steps

Find the tangent point:

(4, 15)

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 = 4 : m = 8

Find the line with slope m=

8

and passing through

(4, 15) : y = 8 x - 17

y = 8 x - 17

y = 2 x^{3} - 5 x, at x = (3 - 2)

\frac{d}{d x} ((3 + x - 2 x^{2})^{\frac{1}{4}})

y^{^{'}} = \frac{( 2 - y ) ^{2}}{2 \cdot 1 + x}

x = 2 y^{2}, x = 4 + y^{2}

y = x^{3}, y = 9 x

What is the tangent of f(x)=x^2-1,\at x=4 ?
The tangent of f(x)=x^2-1,\at x=4 is y=8x-17