What is the tangent of f(x)=sqrt(x),\at x=64.2 ?

The tangent of f(x)=sqrt(x),\at x=64.2 is y=0.06240…x+4.00624…

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tangent of f(x)=sqrt(x),\at x=64.2

tangent of

f (x) = x, at x = 64.2

y = 0.06240 \dots x + 4.00624 \dots

Solution steps

Find the tangent point:

(64.2, 64.2)

Find the slope of

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

Compute the slope of

f (x) = x

x = 64.2 : m = 0.06240 \dots

Find the line with slope m=

0.06240 \dots

and passing through

(64.2, 64.2) : y = 0.06240 \dots x + 4.00624 \dots

y = 0.06240 \dots x + 4.00624 \dots

x \to 3 lim (6 x^{2} + 2)

\int \frac{( x + 1 )}{x ^{2} + 2 x - 4} d x

\int_{1}^{100} \frac{25}{x ^{3}} d x

\int_{0}^{1} 3^{2 x} d x

\int_{0}^{x^{3}} y e^{- \frac{y}{x}} d y

What is the tangent of f(x)=sqrt(x),\at x=64.2 ?
The tangent of f(x)=sqrt(x),\at x=64.2 is y=0.06240…x+4.00624…