Math / Calculus

QuestionSuppose $(1,1)$ is a critical point of a function $f$ with continuous second derivatives. In each case, what can you say about $f$ ? (a) $f_{x x}(1,1)=6, f_{x y}(1,1)=2, f_{y y}(1,1)=1$ The critical point $(1,1)$ is a local minimum. The critical point $(1,1)$ is a local maximum. The critical point $(1,1)$ is a saddle point. Nothing can be determined about the critical point $(1,1)$ . (b) $f_{x x}(1,1)=6, f_{x y}(1,1)=3, f_{y y}(1,1)=1$ The critical point $(1,1)$ is a local minimum. The critical point $(1,1)$ is a local maximum. The critical point $(1,1)$ is a saddle point. Nothing can be determined about the critical point $(1,1)$ .

Studdy Solution

STEP 1

What is this asking? We're figuring out if the point $(1,1)$ is a minimum, maximum, saddle point, or if we just can't tell, based on the second derivatives of the function $f$ at that point! Watch out! Don't forget the second derivatives test formula!

STEP 2

1. Apply the Second Derivatives Test to (a)
2. Apply the Second Derivatives Test to (b)

STEP 3

Let's calculate $D$ at $(1,1)$ using the given values.
Remember, $D = f_{xx}(x,y) \cdot f_{yy}(x,y) - [f_{xy}(x,y)]^2$ .
This helps us classify critical points!

STEP 4

Plugging in our values, we get $D(1,1) = f_{xx}(1,1) \cdot f_{yy}(1,1) - [f_{xy}(1,1)]^2 = 6 \cdot 1 - 2^2 = 6 - 4 = 2$ .
Since $D(1,1)$ is positive, the point $(1,1)$ is either a local minimum or a local maximum.

STEP 5

Since $f_{xx}(1,1) = 6 > 0$ , the function is concave up at $(1,1)$ .
Combining this with the positive $D$ value, we conclude that $(1,1)$ is a local minimum!

STEP 6

Let's calculate $D$ at $(1,1)$ again, this time with the new values.

STEP 7

We have $D(1,1) = f_{xx}(1,1) \cdot f_{yy}(1,1) - [f_{xy}(1,1)]^2 = 6 \cdot 1 - 3^2 = 6 - 9 = -3$ .
Since $D(1,1)$ is negative, the point $(1,1)$ is a saddle point!

STEP 8

(a) The critical point $(1,1)$ is a local minimum. (b) The critical point $(1,1)$ is a saddle point.

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Studdy Solution

STEP 1

What is this asking? We're figuring out if the point $(1,1)$ is a minimum, maximum, saddle point, or if we just can't tell, based on the second derivatives of the function $f$ at that point! Watch out! Don't forget the second derivatives test formula!

STEP 2

1. Apply the Second Derivatives Test to (a)
2. Apply the Second Derivatives Test to (b)

STEP 3

Let's calculate $D$ at $(1,1)$ using the given values.
Remember, $D = f_{xx}(x,y) \cdot f_{yy}(x,y) - [f_{xy}(x,y)]^2$ .
This helps us classify critical points!

STEP 4

Plugging in our values, we get $D(1,1) = f_{xx}(1,1) \cdot f_{yy}(1,1) - [f_{xy}(1,1)]^2 = 6 \cdot 1 - 2^2 = 6 - 4 = 2$ .
Since $D(1,1)$ is positive, the point $(1,1)$ is either a local minimum or a local maximum.

STEP 5

Since $f_{xx}(1,1) = 6 > 0$ , the function is concave up at $(1,1)$ .
Combining this with the positive $D$ value, we conclude that $(1,1)$ is a local minimum!

STEP 6

Let's calculate $D$ at $(1,1)$ again, this time with the new values.

STEP 7

We have $D(1,1) = f_{xx}(1,1) \cdot f_{yy}(1,1) - [f_{xy}(1,1)]^2 = 6 \cdot 1 - 3^2 = 6 - 9 = -3$ .
Since $D(1,1)$ is negative, the point $(1,1)$ is a saddle point!

STEP 8

(a) The critical point $(1,1)$ is a local minimum. (b) The critical point $(1,1)$ is a saddle point.

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Studdy Solution

STEP 1

What is this asking? We're figuring out if the point (1,1) (1,1) (1,1) is a minimum, maximum, saddle point, or if we just can't tell, based on the second derivatives of the function f f f at that point! Watch out! Don't forget the *second derivatives test* formula!

STEP 2

1. Apply the Second Derivatives Test to (a)2. Apply the Second Derivatives Test to (b)

STEP 3

Let's **calculate** D D D at (1,1) (1,1) (1,1) using the given values.Remember, D=fxx(x,y)⋅fyy(x,y)−[fxy(x,y)]2 D = f_{xx}(x,y) \cdot f_{yy}(x,y) - [f_{xy}(x,y)]^2 D=fxx​(x,y)⋅fyy​(x,y)−[fxy​(x,y)]2.This helps us **classify** critical points!

STEP 4

STEP 5

Since fxx(1,1)=6>0 f_{xx}(1,1) = 6 > 0 fxx​(1,1)=6>0, the function is **concave up** at (1,1) (1,1) (1,1).Combining this with the **positive** D D D value, we conclude that (1,1) (1,1) (1,1) is a **local minimum**!

STEP 6

Let's **calculate** D D D at (1,1) (1,1) (1,1) again, this time with the new values.

STEP 7

STEP 8

(a) The critical point (1,1) (1,1) (1,1) is a **local minimum**. (b) The critical point (1,1) (1,1) (1,1) is a **saddle point**.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We're figuring out if the point $(1,1)$ is a minimum, maximum, saddle point, or if we just can't tell, based on the second derivatives of the function $f$ at that point! Watch out! Don't forget the second derivatives test formula!

1. Apply the Second Derivatives Test to (a)
2. Apply the Second Derivatives Test to (b)

Let's calculate $D$ at $(1,1)$ using the given values.
Remember, $D = f_{xx}(x,y) \cdot f_{yy}(x,y) - [f_{xy}(x,y)]^2$ .
This helps us classify critical points!

Since $f_{xx}(1,1) = 6 > 0$ , the function is concave up at $(1,1)$ .
Combining this with the positive $D$ value, we conclude that $(1,1)$ is a local minimum!

Let's calculate $D$ at $(1,1)$ again, this time with the new values.

(a) The critical point $(1,1)$ is a local minimum. (b) The critical point $(1,1)$ is a saddle point.