A stationary point of fx-16 -x2 is

The correct option is A (0, 4) f(x)=√(16 x^2) y=√(16 x^2) On a starting point dydx=0 then dydx=12 2x√(16 x^2)=x√(16 x^2)=0 At x=0 ...

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Range of Quadratic Expression

A stationary ...

Question

A stationary point of $f (x) = \sqrt{16 - x^{2}}$ is

(4, 0)

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(-4, 0)

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(0, 4)

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(-4, 4)

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}}, & i f & x < 0 a, & i f & x = 0 \frac{\sqrt{x}}{\sqrt{16 + \sqrt{4}} - 4}, & i f & x > 0 \end{matrix}

a, f

x = 0

k

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}} f o r x < 0 k f o r x = 0 \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4} f o r x > 0 \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ at x = 0

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}}, x < 0 a, x = 0 \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, x > 0 \end{matrix}

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}}, & w h e n x < 0 a, & w h e n x = 0 \frac{\sqrt{x}}{\sqrt{(16 + \sqrt{x})} - 4}, & w h e n x > 0 \end{matrix}

x = 0

a

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}}, & x < 0 a, & x = 0 \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, & x > 0 \end{matrix}

x = 0

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