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

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

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

tangent of

f (x) = x (x - 2), 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) = x (x - 2) : \frac{df ( x )}{d x} = \frac{3 x - 2}{2 x}

Compute the slope of

f (x) = x (x - 2)

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}

\int_{0}^{2} x^{3} e^{x} d x

x \to 4 - lim (\frac{x - 9}{x - 4})

y^{^{'}} = a - b y

\int t 1 + t t d t

f (x) = 1 - 10 x, at x = - 1

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