What is the tangent of f(x)=sqrt(x^2+55),\at x=3 ?

The tangent of f(x)=sqrt(x^2+55),\at x=3 is y= 3/8 x+55/8

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tangent of f(x)=sqrt(x^2+55),\at x=3

tangent of

f (x) = x^{2} + 55, at x = 3

y = \frac{3}{8} x + \frac{55}{8}

Solution steps

Find the tangent point:

(3, 8)

Find the slope of

f (x) = x^{2} + 55 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 55}

Compute the slope of

f (x) = x^{2} + 55

x = 3 : m = \frac{3}{8}

Find the line with slope m=

\frac{3}{8}

and passing through

(3, 8) : y = \frac{3}{8} x + \frac{55}{8}

y = \frac{3}{8} x + \frac{55}{8}

f (x) = lo g_{4} (x)

x \to 0 lim (\frac{( x + h ) ^{- 2} - h ^{- 2}}{x})

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

\frac{d x}{d y} = x - x^{2}

\int t (t - 2) d t

What is the tangent of f(x)=sqrt(x^2+55),\at x=3 ?
The tangent of f(x)=sqrt(x^2+55),\at x=3 is y= 3/8 x+55/8