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

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

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

tangent of

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

y = x + \frac{3}{4}

Solution steps

Find the tangent point:

(- \frac{1}{2}, \frac{1}{4})

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 1

Find the line with slope m=

1

and passing through

(- \frac{1}{2}, \frac{1}{4}) : y = x + \frac{3}{4}

y = x + \frac{3}{4}

x \to + (- 5) lim (3 x - ∣ x + 5 ∣)

1 - 3 x^{2} y^{^{'}} = x

\frac{\partial}{\partial y} (e^{x^{2} - y^{2} + 4 y})

x \to 1 lim (\frac{x ^{3} - x}{1 - x})

\frac{d}{d x} ((x^{2} + 2 y^{2} + 3 z^{2}) e^{- (x^{2} + y^{2} + z^{2})})

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