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

The tangent of f(x)=(1-2x)(1+2x),\at x=2 is y=-16x+17

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

tangent of

f (x) = (1 - 2 x) (1 + 2 x), at x = 2

y = - 16 x + 17

Solution steps

Find the tangent point:

(2, - 15)

Find the slope of

f (x) = (1 - 2 x) (1 + 2 x) : \frac{df ( x )}{d x} = - 8 x

Compute the slope of

f (x) = (1 - 2 x) (1 + 2 x)

x = 2 : m = - 16

Find the line with slope m=

- 16

and passing through

(2, - 15) : y = - 16 x + 17

y = - 16 x + 17

n = 1 \sum \infty (\frac{3}{8})^{n - 1}

y = \frac{x ^{2} - 2}{2 x + 1}

\int_{0}^{2} x - x^{2} d x

\int sin^{5} (θ) cos (θ) d θ

x^{2} y^{2} = 16, (1, 4)

What is the tangent of f(x)=(1-2x)(1+2x),\at x=2 ?
The tangent of f(x)=(1-2x)(1+2x),\at x=2 is y=-16x+17