What is the tangent of f(x)=|x|sqrt(5-x^2),(2,2) ?

The tangent of f(x)=|x|sqrt(5-x^2),(2,2) is y=-3x+8

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tangent of f(x)=|x|sqrt(5-x^2),(2,2)

tangent of

f (x) = ∣ x ∣ 5 - x^{2}, (2, 2)

y = - 3 x + 8

Solution steps

Find the slope of

f (x) = ∣ x ∣ 5 - x^{2} : \frac{df ( x )}{d x} = \frac{5 x - 2 x ^{3}}{∣ x ∣ - x ^{2} + 5}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 3

Find the line with slope m=

- 3

and passing through

(2, 2) : y = - 3 x + 8

y = - 3 x + 8

x \to 2 lim (3 x + 4 x)

\int_{2}^{4} \frac{2}{x x ^{2} - 4} d x

x \to - 1 + lim (\frac{2}{x + 1})

f (x) = 5 cos (x), at x = \frac{π}{6}

\int \frac{sin ( 5 x )}{cos ^{2} ( 5 x )} d x

What is the tangent of f(x)=|x|sqrt(5-x^2),(2,2) ?
The tangent of f(x)=|x|sqrt(5-x^2),(2,2) is y=-3x+8