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

The tangent of f(x)=(2+4x)(5-4x),\at x=1 is y=-20x+26

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tangent of f(x)=(2+4x)(5-4x),\at x=1

tangent of

f (x) = (2 + 4 x) (5 - 4 x), at x = 1

y = - 20 x + 26

Solution steps

Find the tangent point:

(1, 6)

Find the slope of

f (x) = (2 + 4 x) (5 - 4 x) : \frac{df ( x )}{d x} = - 32 x + 12

Compute the slope of

f (x) = (2 + 4 x) (5 - 4 x)

x = 1 : m = - 20

Find the line with slope m=

- 20

and passing through

(1, 6) : y = - 20 x + 26

y = - 20 x + 26

y = 2 x^{3} - 2 x^{2} - 2 x

\frac{d y}{d x} + \frac{y}{x} = x^{3}

\frac{d}{d x} (\frac{e ^{a x^{2} + b x + c}}{x})

e \to 1 lim (5 ln^{2} (e))

x \to + 0 lim ((1 - \frac{1}{x}) x)

What is the tangent of f(x)=(2+4x)(5-4x),\at x=1 ?
The tangent of f(x)=(2+4x)(5-4x),\at x=1 is y=-20x+26