What is the tangent of y=2sqrt(x)-x+1,(4,1) ?

The tangent of y=2sqrt(x)-x+1,(4,1) is y=-1/2 x+3

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tangent of y=2sqrt(x)-x+1,(4,1)

tangent of

y = 2 x - x + 1, (4, 1)

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

Solution steps

Find the slope of

y = 2 x - x + 1 : \frac{d y}{d x} = \frac{1}{x} - 1

Compute the slope of

y = 2 x - x + 1

(4, 1) : m = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

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

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

y = x e^{- x^{2}}

\int \frac{x + 3}{x - 2} d x

\frac{t ^{2}}{t - 2}

\int (6 x^{5} + 7 x) d x

\int x - x^{2} + x d x

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