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

The tangent of f(x)=(x^2+2)^2,(0,4) is y=4

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tangent of f(x)=(x^2+2)^2,(0,4)

tangent of

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

y = 4

Solution steps

Find the slope of

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

Compute the slope of

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

(0, 4) : m = 0

Find the line with slope m=

0

and passing through

(0, 4) : y = 4

y = 4

x \to 2 lim (\frac{x ^{2} - 4}{x + 4})

\int (3 x - 2) e^{- x} d x

\int 125

f (x) = x^{3} - 12 x + 2

\int_{- 1}^{2} (5 x^{2} + \frac{1}{3} x - \frac{1}{2}) d x

What is the tangent of f(x)=(x^2+2)^2,(0,4) ?
The tangent of f(x)=(x^2+2)^2,(0,4) is y=4