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Let
f be a twice differentiable function, and let
f(1)=-7,
f^(')(1)=0, and
f^('')(1)=-2.
What occurs in the graph of
f at the point
(1,-7) ?
Choose 1 answer:
(A)
(1,-7) is a minimum point.
(B)
(1,-7) is a maximum point.
(C) There's not enough information to tell.

Let $f$ be a twice differentiable function, and let $f(1)=-7$ , $f^{\prime}(1)=0$ , and $f^{\prime \prime}(1)=-2$ . $\newline$ What occurs in the graph of $f$ at the point $(1,-7)$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $(1,-7)$ is a minimum point. $\newline$ (B) $(1,-7)$ is a maximum point. $\newline$ (C) There's not enough information to tell.

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Full solution

Q. Let

f

be a twice differentiable function, and let

f(1)=-7

,

f^{\prime}(1)=0

, and

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

.

\newline

What occurs in the graph of

f

at the point

(1,-7)

?

\newline

Choose

1

answer:

\newline

(A)

(1,-7)

is a minimum point.

\newline

(B)

(1,-7)

is a maximum point.

\newline

(C) There's not enough information to tell.

Analyze Function's Derivatives: To determine the nature of the point $(1,-7)$ on the graph of the function $f$ , we need to analyze the given information about the function's first and second derivatives at the point $x = 1$ .
First Derivative at $x = 1$ : The first derivative of the function $f$ at $x = 1$ is given as $f'(1) = 0$ . This indicates that the slope of the tangent to the graph of $f$ at $x = 1$ is zero, which means the graph has a horizontal tangent line at this point. This could be indicative of a local maximum, local minimum, or a point of inflection.
Second Derivative at $x = 1$ : The second derivative of the function $f$ at $x = 1$ is given as $f''(1) = -2$ . Since the second derivative is negative, it tells us that the graph of $f$ is concave down at $x = 1$ . This implies that the point $(1,-7)$ is a local maximum.
Conclusion: Given the information that $f'(1) = 0$ and $f''(1) = -2$ , we can conclude that the point $(1,-7)$ is a local maximum point on the graph of the function $f$ .

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