Question

Find the minimum and maximum values of the function $y = (t - 3t^2)^{1/3}$ on the interval $[-1, 4]$ by comparing values at the critical points and endpoints. (Use decimal notation. Give your answers to three decimal places.) minimum: maximum:

          Find the minimum and maximum values of the function $y = (t - 3t^2)^{1/3}$ on the interval $[-1, 4]$ by comparing values at the critical points and endpoints.

(Use decimal notation. Give your answers to three decimal places.)
minimum:
maximum:

Find the minimum and maximum values of the function y = (t - 3t^2)^1/3 on the interval [-1, 4] by comparing values at the critical points and endpoints.

(Use decimal notation. Give your answers to three decimal places.)
minimum:
maximum:

Added by Sebastian S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Jonathon Brumley

07/14/2021

Step 1

The derivative of y with respect to t is: y' = (1/3)(t-3t^2)^(-2/3)(1-6t) Setting y' equal to 0 and solving for t: 0 = (1/3)(t-3t^2)^(-2/3)(1-6t) 0 = 1-6t 6t = 1 t = 1/6 Show more…

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Find the minimum and maximum values of the function y=(t-3t^(2))^((1)/(3)) on the interval -1,4 by comparing values at the critical points and endpoints. (Use decimal notation. Give your answers to three decimal places.) minimum: maximum: Find the minimum and maximum values of the function y = (t -- 3t2) on the interval [-1, 4] by comparing values at the critical points and endpoints. (Use decimal notation. Give your answers to three decimal places.) minimum: maximum: