What is the tangent of f(x)=1+sqrt(4+x^2),\at x=0 ?

The tangent of f(x)=1+sqrt(4+x^2),\at x=0 is y=3

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tangent of f(x)=1+sqrt(4+x^2),\at x=0

tangent of

f (x) = 1 + 4 + x^{2}, at x = 0

y = 3

Solution steps

Find the tangent point:

(0, 3)

Find the slope of

f (x) = 1 + 4 + x^{2} : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 4}

Compute the slope of

f (x) = 1 + 4 + x^{2}

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, 3) : y = 3

y = 3

f (x) = 8 x^{2}

f (2)

\frac{1}{2 9}

\int \frac{x ^{3}}{( x ^{4} + 1 )} d x

\int \frac{1}{( x ^{2} - 9 ) ^{2}} d x

What is the tangent of f(x)=1+sqrt(4+x^2),\at x=0 ?
The tangent of f(x)=1+sqrt(4+x^2),\at x=0 is y=3