Let h be a twice differentiable function, and let h(-2)=2, h^(')(-2)=0, and h^('')(-2)=-1. What occurs in the graph of h at the point (-2,2) ? Choose 1 answer: (A) (-2,2) is a minimum point. (B) (-2,2) is a maximum point. (C) There's not enough information to tell.

Analyze Derivatives at -2: To determine what occurs at the point (-2,2), we need to analyze the first and second derivatives of h at x = -2. Given h'(-2) = 0, this means the slope of the tangent line to the graph of h at x = -2 is zero. This indicates a potential maximum, minimum, or inflection point.Check First Derivative: Next, we look at the second derivative, h''(-2) = -1. Since the second derivative is negative, this tells us that the graph of h is concave down at x = -2.Check Second Derivative: A point where the first derivative is zero and the second derivative is negative indicates a local maximum. Therefore, the point (-2,2) is a maximum point on the graph of h.

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Let
h be a twice differentiable function, and let
h(-2)=2,
h^(')(-2)=0, and
h^('')(-2)=-1.
What occurs in the graph of
h at the point
(-2,2) ?
Choose 1 answer:
(A)
(-2,2) is a minimum point.
(B)
(-2,2) is a maximum point.
(C) There's not enough information to tell.

Let $h$ be a twice differentiable function, and let $h(-2)=2$ , $h^{\prime}(-2)=0$ , and $h^{\prime \prime}(-2)=-1$ . $\newline$ What occurs in the graph of $h$ at the point $(-2,2)$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $(-2,2)$ is a minimum point. $\newline$ (B) $(-2,2)$ is a maximum point. $\newline$ (C) There's not enough information to tell.

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Algebra 2

Conjugate root theorems

Full solution

Q. Let

h

be a twice differentiable function, and let

h(-2)=2

h^{\prime}(-2)=0

, and

h^{\prime \prime}(-2)=-1

\newline

What occurs in the graph of

h

at the point

(-2,2)

\newline

Choose

1

answer:

\newline

(A)

(-2,2)

is a minimum point.

\newline

(B)

(-2,2)

is a maximum point.

\newline

Analyze Derivatives at $-2$ : To determine what occurs at the point $(-2,2)$ , we need to analyze the first and second derivatives of $h$ at $x = -2$ . Given $h'(-2) = 0$ , this means the slope of the tangent line to the graph of $h$ at $x = -2$ is zero. This indicates a potential maximum, minimum, or inflection point.
Check First Derivative: Next, we look at the second derivative, $h''(-2) = -1$ . Since the second derivative is negative, this tells us that the graph of $h$ is concave down at $x = -2$ .
Check Second Derivative: A point where the first derivative is zero and the second derivative is negative indicates a local maximum. Therefore, the point $(-2,2)$ is a maximum point on the graph of $h$ .