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

The tangent of f(x)=(1+4x)(5-2x),\at x=2 is y=-14x+37

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

tangent of

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

y = - 14 x + 37

Solution steps

Find the tangent point:

(2, 9)

Find the slope of

f (x) = (1 + 4 x) (5 - 2 x) : \frac{df ( x )}{d x} = - 16 x + 18

Compute the slope of

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

x = 2 : m = - 14

Find the line with slope m=

- 14

and passing through

(2, 9) : y = - 14 x + 37

y = - 14 x + 37

\frac{\partial}{\partial y} (x^{2} + 3 y^{2})

f (x) = sin (sin (x)), 0, at x = π

\int \frac{( x ^{2} )}{x ^{2} - 1} d x

\frac{\partial}{\partial x} (2 x^{3} + x y^{2} + 5 x^{2} + y^{2})

arccos (4 x^{2})

What is the tangent of f(x)=(1+4x)(5-2x),\at x=2 ?
The tangent of f(x)=(1+4x)(5-2x),\at x=2 is y=-14x+37