Find the set of x values where f(x) has relative minima and maxima

Question

A) f(x)=3x^2-15x-10

1) Find set of critical points of f(x)

2)find the intervals on which f(x) is increasing

3)find the intervals on which f(x) is decreasing

4) find the set of x-values where f(x) has relative minima

5) find the set of x-values where f(x) has relative maxima

B) f(x)= (x^2+2)/(2x^2+1)

1) Find set of critical points of f(x)

2)find the intervals on which f(x) is increasing

3)find the intervals on which f(x) is decreasing

4) find the set of x-values where f(x) has relative minima

5) find the set of x-values where f(x) has relative maxima

Answer 1

A-(1)

The function is .

Differentiate with respect to x.

To find the critical points, equate to zero.

Critical number is .

Substitute in the function.

Therefore, the critical point is .

Answer 2

A-(2)&(3)

The function is .

From A-(1), the critical number is .

The derivative function is .

Consider test intervals as and .

Therefore the function is increasing in the interval .

The function is decreasing in the interval .

Solution:

A-(2): The function is increasing in the interval .

A-(3): The function is decreasing in the interval .

Answer 3

A-(4)&(5)

The function is .

From A-(1), the critical number is .

The derivative function is .

Differentiate with respect to x.

The sign of is positive for all values of x.

Therefore, the function has relative minimum at .

Relative minima is .

No relative maxima.

Solution:

A-(4): Relative minima is .

A-(5): No relative maxima.

Answer 4

B-(1)

The function is .

Differentiate with respect to .

.

To find the critical points, equate to zero.

Critical number is .

Substitute  in the function.

Therefore, the critical point is .

Answer 5

B-(2)&(3)

The function is .

From B-(1), the critical point is .

The derivative function is .

Consider test intervals as and .

Therefore, the function is increasing in the interval .

The function is decreasing in the interval .

Solution:

B-(2): The function is increasing in the interval .

B-(3): The function is decreasing in the interval .

Answer 6

B-(4)&(5)

The function is .

From B-(1), the critical number is .

The derivative function is .

Differentiate with respect to x.

.

Find the value of second derivative at .

.

Therefore, the function has relative maximum at .

Relative maxima is .

No relative minima.

Solution:

B-(4): Relative minima is .

B-(5): No relative minima.