Absolute Extremas of f(x) = x^4 - 10x^2 - 5 on intervals

[mrderson]

I have a math problem where i need to find the absolute maximums and minimums but i only found the maximums by using the table from the graph on my calculator. the question is...

Find the absolute maximum and absolute minimum values of the function f(x) = x^4 - 10 x^2 - 5 on each of the indicated intervals. Enter None for any absolute extrema that do not exist.

(A) Interval = [-3,-1].
Absolute maximum = -14
Absolute minimum =

(B) Interval = [-4,1].
Absolute maximum = 91
Absolute minimum =

(C) Interval = [-3,4].
Absolute maximum = 91
Absolute minimum =

I thought to find the mins and maxs you had to use the derivative but i did something wrong

 

[stapel]

I thought to find the mins and maxs you had to use the derivative but i did something wrong

You do use the derivative to find max/min points. What did you do? What were your steps? What values did you get? (We can't find errors, if any, in work we can't see.)

Please be complete. Thank you!

 

[soroban]

Hello, mrderson!

Find the absolute maximum and absolute minimum values of: .\(\displaystyle f(x) \:=\: x^4 - 10 x^2 - 5\)

(B) on the Interval [-4,1]
Absolute maximum = 91
Absolute minimum = . . . . How did you miss the minimum?

\(\displaystyle \text{Find the critical points: }\:\text{set }f'(x) = 0\,\text{ and solve.}\)

. . \(\displaystyle f'(x) \:=\:4x^3-20x \:=\:0 \quad\Rightarrow\quad 4x(x^2-5) \:=\:0\)
. . \(\displaystyle \text{Hence: }\;x \:=\:\text{-}\sqrt{5},\;0,\:\rlap{///}\sqrt{5}\)
. . . . . . . . . . . . .\(\displaystyle \overbrace{^{\text{not on the interval}}}\)


\(\displaystyle \text{Second derivative test: }\:f''(x) \:=\:12x^2-20\)

. . \(\displaystyle f''(\text{-}\sqrt{5}) \,=\,+40\quad\text{concave up: minimum}\)

. . \(\displaystyle f''(0) \,=\,-20\quad\text{concave down: maximum}\)


\(\displaystyle \text{Therefore: }\;\begin{array}{ccccc}f(\text{-}4) &=&91 & & \text{abs. max} \\ f(\text{-}\sqrt{5}) &=& \text{-}30 & & \text{abs. min} \\ f(0) &=& \text{-}5 & & \text{rel. max} \\ f(1) &=& \text{-}14 & & \text{rel. min} \end{array}\)