What are the extrema of $f (x) = 1 - \sqrt{x}$ $f(x) = 1 - sqrt(x)$ ?

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Answer 1 · 2016-12-08T10:51:45

Max f = 1. There is no minimum.

$y = f (x) = 1 - \sqrt{x}$ $y=f(x)=1-sqrtx$ . Graph is inserted.

This represents a semi parabola, in the quadrants $Q_{1} and Q_{4}$ $Q_1 and Q_4$ ,

wherein x >=0.

Max y is at the end (0, 1). Of course, there is no minimum.

Note that, as $x \to \infty, y \to - \infty$ $x to oo, y to -oo$ .

The parent equation is ${(y - 1)}^{2} = x$ $(y-1)^2 = x$ that can be separated into

$y = 1 \pm \sqrt{x}$ $y=1+-sqrtx$ .

graph{y+sqrtx-1=0 [-2.5, 2.5, -1.25, 1.25]}

What are the extrema of f(x)=1−√xf(x) = 1 - sqrt(x)?