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}

Compute the slope of

f (x) = x x

x = 9 : 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}

\int cos^{(5)} (5 x) sin (5 x) d x

\int_{0}^{9} 3 x - x d x

\int \frac{x}{( x - 1 ) ^{2} + 1} d x

\int x e^{3} d x

- \frac{48 x}{( x ^{2} + 12 ) ^{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