What is the tangent of f(x)=4-x^2,\at x=2 ?

The tangent of f(x)=4-x^2,\at x=2 is y=-4x+8

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tangent of f(x)=4-x^2,\at x=2

tangent of

f (x) = 4 - x^{2}, at x = 2

y = - 4 x + 8

Solution steps

Find the tangent point:

(2, 0)

Find the slope of

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

Compute the slope of

f (x) = 4 - x^{2}

x = 2 : m = - 4

Find the line with slope m=

- 4

and passing through

(2, 0) : y = - 4 x + 8

y = - 4 x + 8

\int cos (\frac{1}{2} x) d x

\frac{d}{d x} (\frac{x ^{5}}{5 !})

\frac{d}{d x} (x^{2} + cos (x))

n = 1 \sum \infty \frac{4 ^{n}}{4 ^{n}}

\frac{d}{d x} (4 x^{4} e^{x})

What is the tangent of f(x)=4-x^2,\at x=2 ?
The tangent of f(x)=4-x^2,\at x=2 is y=-4x+8