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

The tangent of f(x)=(x^2+3x-5)(1-x),\at x=2 is y=-12x+19

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

tangent of

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

y = - 12 x + 19

Solution steps

Find the tangent point:

(2, - 5)

Find the slope of

f (x) = (x^{2} + 3 x - 5) (1 - x) : \frac{df ( x )}{d x} = - 3 x^{2} - 4 x + 8

Compute the slope of

f (x) = (x^{2} + 3 x - 5) (1 - x)

x = 2 : m = - 12

Find the line with slope m=

- 12

and passing through

(2, - 5) : y = - 12 x + 19

y = - 12 x + 19

\frac{3 x}{x + 1}

\int \frac{1}{2 x ^{2} - 3} d x

\int_{0}^{1} (x^{2} + 2 x + 3) e^{2 x} d x

y = ln (x^{4} (x))

f (x) = 3 x - x^{2}

What is the tangent of f(x)=(x^2+3x-5)(1-x),\at x=2 ?
The tangent of f(x)=(x^2+3x-5)(1-x),\at x=2 is y=-12x+19