What is the tangent of f(x)=xsqrt(x),\at x=36 ?

The tangent of f(x)=xsqrt(x),\at x=36 is y=9x-108

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tangent of f(x)=xsqrt(x),\at x=36

tangent of

f (x) = x x, at x = 36

y = 9 x - 108

Solution steps

Find the tangent point:

(36, 216)

Find the slope of

f (x) = x x : \frac{df ( x )}{d x} = \frac{3 x}{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 = 9

Find the line with slope m=

9

and passing through

(36, 216) : y = 9 x - 108

y = 9 x - 108

What is the tangent of f(x)=xsqrt(x),\at x=36 ?
The tangent of f(x)=xsqrt(x),\at x=36 is y=9x-108