{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw12solutions

# hw12solutions - Math 2250 Written HW#12 Solutions 1...

This preview shows pages 1–2. Sign up to view the full content.

Math 2250 Written HW #12 Solutions 1. Consider f ( x ) = 1 4 x 2 + 7 (a) Find all critical points of f ( x ). For each one determine whether it is a local maximum, a local minimum, or neither. Answer: To find the critical points, first we need to compute f 0 ( x ) = - 1 (4 x 2 + 7) 2 · 8 x = - 8 x (4 x 2 + 7) 2 . Since the denominator is always positive, f 0 ( x ) = 0 exactly when the numerator is zero, so the only critical point occurs when x = 0, so this is the point (0 , 1 / 7). Now, again since the denominator is always positive, f 0 ( x ) has the same sign as the numerator, which is - 8 x . Therefore, f 0 ( x ) changes sign from positive to negative at x = 0, so the first derivative test implies that (0 , 1 / 7) is a local maximum. (b) If it exists, what is the absolute maximum of f ( x )? If it exists, what is the absolute minimum? Answer: Since f ( x ) has only one critical point, and that critical point is a local max- imum, it must also be an absolute maximum. Therefore, the absolute maximum value of f ( x ) is 1 / 7. On the other hand, any absolute minimum would also have to be a local

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}