What is the tangent of f(x)= 3/(x^{1/2)},(1,3) ?

The tangent of f(x)= 3/(x^{1/2)},(1,3) is y=-3/2 x+9/2

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tangent of f(x)= 3/(x^{1/2)},(1,3)

tangent of

f (x) = \frac{3}{x ^{\frac{1}{2}}}, (1, 3)

y = - \frac{3}{2} x + \frac{9}{2}

Solution steps

Find the slope of

f (x) = \frac{3}{x ^{\frac{1}{2}}} : \frac{df ( x )}{d x} = - \frac{3}{2 x ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{3}{x ^{\frac{1}{2}}}

(1, 3) : m = - \frac{3}{2}

Find the line with slope m=

- \frac{3}{2}

and passing through

(1, 3) : y = - \frac{3}{2} x + \frac{9}{2}

y = - \frac{3}{2} x + \frac{9}{2}

n = 1 \sum \infty (\frac{4}{5})^{n}

\int_{7}^{8} \frac{7}{50 + x ^{2} - 14 x} d x

\int 2 x^{\frac{5}{2}} d x

x \to 5 lim (f (x))

y = x^{7 x}

What is the tangent of f(x)= 3/(x^{1/2)},(1,3) ?
The tangent of f(x)= 3/(x^{1/2)},(1,3) is y=-3/2 x+9/2