For x ∈ mathbbR, if f(x) = -(x-1)^2 + 2, then (A) f is an increasing function on (-∞, 1] (B) f has no critical points (C) f has a maximum value at x…

The function $f(x) = -(x-1)^2 + 2$f(x)=−(x−1)2+2 is a quadratic function in vertex form $f(x) = a(x-h)^2 + k$f(x)=a(x−h)2+k.

From the given function:

• Vertex: $(h, k) = (1, 2)$(h,k)=(1,2)
• Coefficient: $a = -1 < 0$a=−1<0

Since $a < 0$a<0, the parabola opens downward.

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For a downward-opening parabola with vertex at $x = 1$x=1:

• The function increases on $(-\infty, 1]$(−∞,1]
• The function decreases on $[1, \infty)$[1,∞)

Statement (A) is TRUE.

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To find critical points, take the derivative:

$$f'(x) = -2(x-1)$$f′(x)=−2(x−1)

Setting $f'(x) = 0$f′(x)=0:

$$-2(x-1) = 0$$−2(x−1)=0

$$x = 1$$x=1

A critical point exists at $x = 1$x=1.

Statement (B) is FALSE.

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Since the parabola opens downward, the vertex represents the maximum point.

At $x = 1$x=1:

$$f(1) = -(1-1)^2 + 2$$f(1)=−(1−1)2+2

$$f(1) = 2$$f(1)=2

The function has a maximum value of $2$2 at $x = 1$x=1.

Statement (C) is TRUE.

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Since the parabola opens downward, the vertex represents a maximum, not a minimum.

As $x \to \pm\infty$x→±∞, $f(x) \to -\infty$f(x)→−∞, so no minimum value exists.

Statement (D) is FALSE.

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The correct statements are (A) and (C) only.