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

The tangent of f(x)=2-3x^2,\at x=1 is y=-6x+5

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

tangent of

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

y = - 6 x + 5

Solution steps

Find the tangent point:

(1, - 1)

Find the slope of

f (x) = 2 - 3 x^{2} : \frac{df ( x )}{d x} = - 6 x

Compute the slope of

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

x = 1 : m = - 6

Find the line with slope m=

- 6

and passing through

(1, - 1) : y = - 6 x + 5

y = - 6 x + 5

\frac{d}{d x} ((x^{3})^{2})

\int \frac{1}{a x + b x ^{2}} d x

x \to - 1 lim (\frac{x ^{2} - 1}{x + 1})

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{1 + e ^{x}}

\frac{\partial}{\partial z} (x cos (\frac{7 y}{z}))

What is the tangent of f(x)=2-3x^2,\at x=1 ?
The tangent of f(x)=2-3x^2,\at x=1 is y=-6x+5