What is the absolute maximum for f(x) = 2 + x^2 on [-2, 3]?

Question

Is it

-3 ?

Answer 1

f(x) = x² + 2
f'(x) = 3x

f'(x) = 3x = 0 means that x = 0 is a critical point.
Check this and your endpoints.

f(-2) = (-2)² + 2 = 4 + 2 = 6
f(0) = (0)² + 2 = 0 + 2 = 2 (minimum)
f(3) = (3)² + 2 = 9 + 2 = 11.

The absolute max over your interval is at x = 3, f(x) = 11.

Answer 2

3

Answer 3

Before resorting to tricky calculus, I'd recommend that you examine the equation.

For simple equations like these, the graph of the equation is a U shape that is symmetrical about the y axis.

Thus by logical deduction, it follows that the maximum value the function can take in the given range falls at x=3.

Thus the absolute maximum is f(3) = 2 + 9 = 11

I hope this provides an alternative way of answering the question. After all, I feel maths is all about finding as many ways of solving a question as possible :)

Answer 4

Taking the first and second derivatives of f(x)

f'(x) = 2x
f"(x) = 2

The function has an extreme where x = 0, but it's the minimum of the function. There, compare the endpoints of the range instead.

f(x=-2) = 2 + (-2)^2 = 6
f(x=3) = 2 + (3)^2 = 11

The maximum of the function on the given range is at x = 3 so that f(3) = 11.

Hope that helps.

Answer 5

Take the derrivative
f(x)=2+x^2
f'(x)=2x
Find critical numbers by setting derrivative=0
2x=0
x=0
Test a point to the left and right of the critical number
f'(-1)=-2
f'(1)=2
The derrivative shifts from - to + telling you that a minimum, not a maximum occurs at the point
Since the function is on a closed domain you can plug in the endpoints
f(-2)=6
f(3)=11
By doing this you find that the maximum of the function is 11 occuring at x=3.
This one could also be done easily just by graphing the function and looking at it

Answer 6

put it in your calculator.

Answer 7

1.6