What is the tangent of f(x)= 1/(4x^{3/2)},\at x=4 ?

The tangent of f(x)= 1/(4x^{3/2)},\at x=4 is y=-3/256 x+5/64

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tangent of f(x)= 1/(4x^{3/2)},\at x=4

tangent of

f (x) = \frac{1}{4 x ^{\frac{3}{2}}}, at x = 4

y = - \frac{3}{256} x + \frac{5}{64}

Solution steps

Find the tangent point:

(4, \frac{1}{32})

Find the slope of

f (x) = \frac{1}{4 x ^{\frac{3}{2}}} : \frac{df ( x )}{d x} = - \frac{3}{8 x ^{\frac{5}{2}}}

Compute the slope of

f (x) = \frac{1}{4 x ^{\frac{3}{2}}}

x = 4 : m = - \frac{3}{256}

Find the line with slope m=

- \frac{3}{256}

and passing through

(4, \frac{1}{32}) : y = - \frac{3}{256} x + \frac{5}{64}

y = - \frac{3}{256} x + \frac{5}{64}

\frac{d y}{d x} = 6 x^{2} y

\int e^{r^{2}} d r

x \to 0 lim (\frac{ln ( 1 )}{x})

\int (x^{7} - 2 x^{6})^{5} (7 x^{6} - 12 x^{5}) d x

n = 1 \sum \infty n e^{- 8 n}

What is the tangent of f(x)= 1/(4x^{3/2)},\at x=4 ?
The tangent of f(x)= 1/(4x^{3/2)},\at x=4 is y=-3/256 x+5/64