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

The tangent of f(x)=x^2,(-1,1) is y=-2x-1

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

tangent of

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

y = - 2 x - 1

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x^{2}

(- 1, 1) : m = - 2

Find the line with slope m=

- 2

and passing through

(- 1, 1) : y = - 2 x - 1

y = - 2 x - 1

\int x cos (2 x^{2} + 7) d x

h \to 0 lim (\frac{( 2. 7 ^{h} - 1 )}{h})

\int sin^{- 2} (x) cos^{3} (x) d x

\int 3 t d t

\frac{d}{d x} (12 x^{2} - 2 x^{3})

What is the tangent of f(x)=x^2,(-1,1) ?
The tangent of f(x)=x^2,(-1,1) is y=-2x-1