Studdy AI Logo
Math  /  Calculus

QuestionFind the absolute extrema if they exist, as well as all values of xx where they occur, for the function f(x)=2x−224x27f(x)=2 x-224 x^{\frac{2}{7}} (a) on the interval [−128,122][-128,122] and (b)(b) on the interval [122,384][122,384]. (a) Identify the absolute maximum on the interval [−128,122][-128,122] if it exists. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

Studdy Solution

STEP 1

1. The function f(x)=2x−224x27 f(x) = 2x - 224x^{\frac{2}{7}} is continuous on the given intervals.
2. To find absolute extrema, we need to evaluate the function at critical points and endpoints of the interval.
3. Critical points occur where the derivative is zero or undefined.

STEP 2

1. Find the derivative of the function.
2. Determine critical points by setting the derivative equal to zero.
3. Evaluate the function at critical points and endpoints of the interval [−128,122][-128, 122].
4. Compare these values to determine the absolute maximum on the interval [−128,122][-128, 122].

STEP 3

Find the derivative of the function f(x)=2x−224x27 f(x) = 2x - 224x^{\frac{2}{7}} .
f′(x)=ddx(2x)−ddx(224x27) f'(x) = \frac{d}{dx}(2x) - \frac{d}{dx}(224x^{\frac{2}{7}})
f′(x)=2−224⋅27⋅x27−1 f'(x) = 2 - 224 \cdot \frac{2}{7} \cdot x^{\frac{2}{7} - 1}
f′(x)=2−4487x−57 f'(x) = 2 - \frac{448}{7}x^{-\frac{5}{7}}
f′(x)=2−64x−57 f'(x) = 2 - 64x^{-\frac{5}{7}}

STEP 4

Determine critical points by setting the derivative equal to zero:
2−64x−57=0 2 - 64x^{-\frac{5}{7}} = 0
64x−57=2 64x^{-\frac{5}{7}} = 2
x−57=132 x^{-\frac{5}{7}} = \frac{1}{32}
Raise both sides to the power of −75-\frac{7}{5} to solve for xx:
x=(132)−75 x = \left(\frac{1}{32}\right)^{-\frac{7}{5}}
x=3275 x = 32^{\frac{7}{5}}
Calculate x x :
x=(25)75=27=128 x = (2^5)^{\frac{7}{5}} = 2^7 = 128
Since x=128 x = 128 is not within the interval [−128,122][-128, 122], no critical points exist within this interval.

STEP 5

Evaluate the function at the endpoints of the interval [−128,122][-128, 122]:
1. Evaluate at x=−128 x = -128 :
f(−128)=2(−128)−224(−128)27 f(-128) = 2(-128) - 224(-128)^{\frac{2}{7}}
2. Evaluate at x=122 x = 122 :
f(122)=2(122)−224(122)27 f(122) = 2(122) - 224(122)^{\frac{2}{7}}
Calculate these values:
f(−128)=−256−224×4=−256−896=−1152 f(-128) = -256 - 224 \times 4 = -256 - 896 = -1152
f(122)=244−224×4.57≈244−1023.68≈−779.68 f(122) = 244 - 224 \times 4.57 \approx 244 - 1023.68 \approx -779.68

STEP 6

Compare the function values at the endpoints to determine the absolute maximum on the interval [−128,122][-128, 122]:
- f(−128)=−1152 f(-128) = -1152 - f(122)≈−779.68 f(122) \approx -779.68
The absolute maximum value is ≈−779.68 \approx -779.68 at x=122 x = 122 .
The absolute maximum on the interval [−128,122][-128, 122] is approximately −779.68-779.68 at x=122 x = 122 .

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord