What is the tangent of f(x)=(2+4x)(6-5x),\at x=1 ?

The tangent of f(x)=(2+4x)(6-5x),\at x=1 is y=-26x+32

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tangent of f(x)=(2+4x)(6-5x),\at x=1

tangent of

f (x) = (2 + 4 x) (6 - 5 x), at x = 1

y = - 26 x + 32

Solution steps

Find the tangent point:

(1, 6)

Find the slope of

f (x) = (2 + 4 x) (6 - 5 x) : \frac{df ( x )}{d x} = - 40 x + 14

Compute the slope of

f (x) = (2 + 4 x) (6 - 5 x)

x = 1 : m = - 26

Find the line with slope m=

- 26

and passing through

(1, 6) : y = - 26 x + 32

y = - 26 x + 32

\int_{- 2}^{- 4} (- 4)^{- 3} d x

\int_{- 1}^{1} x d x

f (t) = t^{3} e^{2 t}

\frac{d}{d x} (- 2 + 6 sin (6 x))

ln (x^{2})

What is the tangent of f(x)=(2+4x)(6-5x),\at x=1 ?
The tangent of f(x)=(2+4x)(6-5x),\at x=1 is y=-26x+32