What is the tangent of f(x)=(2+3x)(6-5x),\at x=1 ?

The tangent of f(x)=(2+3x)(6-5x),\at x=1 is y=-22x+27

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tangent of f(x)=(2+3x)(6-5x),\at x=1

tangent of

f (x) = (2 + 3 x) (6 - 5 x), at x = 1

y = - 22 x + 27

Solution steps

Find the tangent point:

(1, 5)

Find the slope of

f (x) = (2 + 3 x) (6 - 5 x) : \frac{df ( x )}{d x} = - 30 x + 8

Compute the slope of

f (x) = (2 + 3 x) (6 - 5 x)

x = 1 : m = - 22

Find the line with slope m=

- 22

and passing through

(1, 5) : y = - 22 x + 27

y = - 22 x + 27

\frac{d}{d x} (e^{(3 - x)})

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\int cos (x) (4 + 4 sin^{2} (x)) d x

x \to 0 lim (\frac{x - x}{x ^{2}})

\frac{d}{d x} ((x + 1) ln^{2} (x + 1))

What is the tangent of f(x)=(2+3x)(6-5x),\at x=1 ?
The tangent of f(x)=(2+3x)(6-5x),\at x=1 is y=-22x+27