What is the tangent of f(x)=xsqrt(x),\at x=18 ?

The tangent of f(x)=xsqrt(x),\at x=18 is y= 9/(sqrt(2))x-27sqrt(2)

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tangent of f(x)=xsqrt(x),\at x=18

tangent of

f (x) = x x, at x = 18

y = \frac{9}{2} x - 272

Solution steps

Find the tangent point:

(18, 542)

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 = 18 : m = \frac{9}{2}

Find the line with slope m=

\frac{9}{2}

and passing through

(18, 542) : y = \frac{9}{2} x - 272

y = \frac{9}{2} x - 272

x \to 2 + lim (\frac{1 - x}{x - 2})

\int_{0}^{2} \frac{1}{4 - x ^{2}} d x

\int \frac{5 x ^{2} + 15 x + 8}{( x + 2 ) ^{2} x} d x

y = e^{- 3 x}

What is the tangent of f(x)=xsqrt(x),\at x=18 ?
The tangent of f(x)=xsqrt(x),\at x=18 is y= 9/(sqrt(2))x-27sqrt(2)