What is the tangent of f(x)=(2x-1)(x+2),\at x=(1.3) ?

The tangent of f(x)=(2x-1)(x+2),\at x=(1.3) is y=8.2x-5.38

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tangent of f(x)=(2x-1)(x+2),\at x=(1.3)

tangent of

f (x) = (2 x - 1) (x + 2), at x = (1.3)

y = 8.2 x - 5.38

Solution steps

Find the tangent point:

(1.3, 5.28)

Find the slope of

f (x) = (2 x - 1) (x + 2) : \frac{df ( x )}{d x} = 4 x + 3

Compute the slope of

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

x = 1.3 : m = 8.2

Find the line with slope m=

8.2

and passing through

(1.3, 5.28) : y = 8.2 x - 5.38

y = 8.2 x - 5.38

\int \frac{ln ( 14 x )}{x ^{6}} d x

ln (t + 1)

\frac{\partial}{\partial x} (e^{- 3 y} cos (- 3 x))

\int \frac{1}{1 + t ^{2}} d t

\int_{1}^{x} 2 (t - 1) d t

What is the tangent of f(x)=(2x-1)(x+2),\at x=(1.3) ?
The tangent of f(x)=(2x-1)(x+2),\at x=(1.3) is y=8.2x-5.38