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

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

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

tangent of

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

y = - x - \frac{1}{2}

Solution steps

Find the tangent point:

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

Find the slope of

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

Compute the slope of

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

x = - 1 : m = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x - \frac{1}{2}

4 e^{3 t}

\int sin (\frac{π}{2} x) d x

x \to - 1 lim (- f (x))

\int_{l n (\frac{5}{6})}^{\infty} \frac{e ^{x}}{36 e ^{2 x} + 25} d x

\frac{d y}{d x} + \frac{3}{x} y = \frac{ln ( x )}{x ^{2}}

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