What is the tangent of f(x)=sqrt(x),\at x=5 ?

The tangent of f(x)=sqrt(x),\at x=5 is y= 1/(2sqrt(5))x+sqrt(5)-(sqrt(5))/2

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tangent of f(x)=sqrt(x),\at x=5

tangent of

f (x) = x, at x = 5

y = \frac{1}{2 5} x + 5 - \frac{5}{2}

Solution steps

Find the tangent point:

(5, 5)

Find the slope of

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

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{1}{2 5}

Find the line with slope m=

\frac{1}{2 5}

and passing through

(5, 5) : y = \frac{1}{2 5} x + 5 - \frac{5}{2}

y = \frac{1}{2 5} x + 5 - \frac{5}{2}

f^{^{''}} (t) - f (t) = - 8 t, f (3) = - 8

f (x) = \frac{2 x - 1}{4 x + 8}

\frac{1}{1 + x ^{3}}

\frac{\partial}{\partial y} (4 y - 2 x)

\int 4 - x^{2} x d x

What is the tangent of f(x)=sqrt(x),\at x=5 ?
The tangent of f(x)=sqrt(x),\at x=5 is y= 1/(2sqrt(5))x+sqrt(5)-(sqrt(5))/2