What is the tangent of f(x)=(6-3x)(6+3x),\at x=3 ?

The tangent of f(x)=(6-3x)(6+3x),\at x=3 is y=-54x+117

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tangent of f(x)=(6-3x)(6+3x),\at x=3

tangent of

f (x) = (6 - 3 x) (6 + 3 x), at x = 3

y = - 54 x + 117

Solution steps

Find the tangent point:

(3, - 45)

Find the slope of

f (x) = (6 - 3 x) (6 + 3 x) : \frac{df ( x )}{d x} = - 18 x

Compute the slope of

f (x) = (6 - 3 x) (6 + 3 x)

x = 3 : m = - 54

Find the line with slope m=

- 54

and passing through

(3, - 45) : y = - 54 x + 117

y = - 54 x + 117

\int_{- 0.5}^{0.5} 1.5 x^{2} d x

y = (3 x - 2)^{2} (5 x + 4)

\frac{- 1}{s ^{2} - 1}

\frac{d}{d x} (\frac{2}{x} + x)

\frac{\partial}{\partial v} (\frac{1}{u ^{2} + v ^{2}})

What is the tangent of f(x)=(6-3x)(6+3x),\at x=3 ?
The tangent of f(x)=(6-3x)(6+3x),\at x=3 is y=-54x+117