What is the tangent of f(x)=xsqrt(x),\at x=9 ?

The tangent of f(x)=xsqrt(x),\at x=9 is y= 9/2 x-27/2

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tangent of f(x)=xsqrt(x),\at x=9

tangent of

f (x) = x x, at x = 9

y = \frac{9}{2} x - \frac{27}{2}

Solution steps

Find the tangent point:

(9, 27)

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 = \frac{9}{2}

Find the line with slope m=

\frac{9}{2}

and passing through

(9, 27) : y = \frac{9}{2} x - \frac{27}{2}

y = \frac{9}{2} x - \frac{27}{2}

What is the tangent of f(x)=xsqrt(x),\at x=9 ?
The tangent of f(x)=xsqrt(x),\at x=9 is y= 9/2 x-27/2