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

The tangent of f(x)=3-x-x^2,\at x=1 is y=-3x+4

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

tangent of

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

y = - 3 x + 4

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

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

Compute the slope of

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

x = 1 : m = - 3

Find the line with slope m=

- 3

and passing through

(1, 1) : y = - 3 x + 4

y = - 3 x + 4

\frac{d}{d x} (x^{3} + 4 x^{2} - 6 x + 1)

20 y^{^{''}} - 11 y^{^{'}} - 3 y = 0

\int 5 sin^{3} (5 x co) s^{3} 5 x d x

\int_{9}^{10} \frac{1}{( x - 9 ) ^{\frac{5}{4}}} d x

\frac{d}{d x} (\frac{x}{2} - tan (x))

What is the tangent of f(x)=3-x-x^2,\at x=1 ?
The tangent of f(x)=3-x-x^2,\at x=1 is y=-3x+4