f(x) = 4x² - 4x + 6 on (-$\infty$, $\infty$) has A) only an absolute maximum B) only an absolute minimum C) both an absolute maximum and an absolute minimum D) neither an absolute maximum nor an absolute minimum. E) cannot be determined

Question

Quizplus · Accepted Answer

We can start by taking the first derivative of f(x):

f'(x) = 4S1U1P12S1S1P0 F1F1F1S1-F1F1F10

Setting f'(x) = 0, we get:

4S1U1P12S1S1P0 F1F1F1S1-F1F1F10 = 0

This equation only has one solution, which is x = -6/4 = -1.5

Now we can take the second derivative of f(x):

f''(x) = 4S1U1P12S1S1P0 F1F1F1S1S1S1P0

Since this expression is always positive, we know that x = -1.5 is a minimum point.

Therefore, f(x) has an absolute minimum at x = -1.5.

Now let's check the behavior of f(x) as x approaches -F1F1F1S1F1F1F10 and F1F1F1S1F1F1F10.

As x approaches -F1F1F1S1F1F1F10, we have:

f(x) = 4xS1U1P12S1S1P0 F1F1F1S1-F1F1F10 + 6

This expression goes to -S1S1, which is not a finite value. Therefore, f(x) does not have an absolute maximum at x = -F1F1F1S1F1F1F10.

As x approaches F1F1F1S1F1F1F10, we have:

f(x) = 4xS1U1P12S1S1P0 F1F1F1S1-F1F1F10 + 6

This expression goes to S1S1, which is not a finite value. Therefore, f(x) does not have an absolute maximum at x = F1F1F1S1F1F1F10.

In conclusion, f(x) has an absolute minimum at x = -1.5, but it does not have an absolute maximum. Therefore, the correct answer is C) both an absolute maximum and an absolute minimum.

F(x) = 4x2 - 4x + 6 on (- ∞\infty∞ ∞\infty∞

F(x) = 4x² - 4x + 6 on (- $\infty$ $\infty$