What is the tangent of f(x)= 1/(sqrt(x)),\at x= 1/16 ?

The tangent of f(x)= 1/(sqrt(x)),\at x= 1/16 is y=-32x+6

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tangent of f(x)= 1/(sqrt(x)),\at x= 1/16

tangent of

f (x) = \frac{1}{x}, at x = \frac{1}{16}

y = - 32 x + 6

Solution steps

Find the tangent point:

(\frac{1}{16}, 4)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 32

Find the line with slope m=

- 32

and passing through

(\frac{1}{16}, 4) : y = - 32 x + 6

y = - 32 x + 6

What is the tangent of f(x)= 1/(sqrt(x)),\at x= 1/16 ?
The tangent of f(x)= 1/(sqrt(x)),\at x= 1/16 is y=-32x+6