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

\int e^{4 x} sin (3 x) d x

x \to 0 + lim (\frac{7}{x} - 7 ∣ x ∣)

\frac{1}{x - 1}, 12

\int \frac{e ^{x}}{e ^{x} + 4} d x

y = x^{3}, x = y^{2}

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