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

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

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

tangent of

f (x) = 2 - x, at x = 1

y = - \frac{1}{2} x + \frac{3}{2}

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

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

Compute the slope of

f (x) = 2 - x

x = 1 : m = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

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

y = - \frac{1}{2} x + \frac{3}{2}

\frac{\partial}{\partial x} (x^{2} \cdot sin (2 y))

2 e^{x} - 5 cos (x) + 3

f (x) = x ln (x)

x \to 2 lim (x^{2} - 6 x)

y^{^{''}} + 5 y^{^{'}} + 4 y = - 18 t e^{4 t}

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