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

The tangent of f(x)=(1+3x)(5-2x),\at x=2 is y=-11x+29

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

tangent of

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

y = - 11 x + 29

Solution steps

Find the tangent point:

(2, 7)

Find the slope of

f (x) = (1 + 3 x) (5 - 2 x) : \frac{df ( x )}{d x} = - 12 x + 13

Compute the slope of

f (x) = (1 + 3 x) (5 - 2 x)

x = 2 : m = - 11

Find the line with slope m=

- 11

and passing through

(2, 7) : y = - 11 x + 29

y = - 11 x + 29

\frac{s}{( s + 1 ) ( s ^{2} + 4 )}

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x ^{10}})

\int - tan (t) d t

\frac{3}{( s + 1 ) ^{2} + 1}

\frac{\partial}{\partial x} (\frac{y}{x} - 1)

What is the tangent of f(x)=(1+3x)(5-2x),\at x=2 ?
The tangent of f(x)=(1+3x)(5-2x),\at x=2 is y=-11x+29