Find the absolute extrema of the function on the closed interval. y=3x23−2x, [−1, 1] minimum - ? maximum - ?

y=3x23−2x, [−1, 1] Now dydx=2x13−2 =2x13−2 =2[1−x13]x13 x=0⇒f(0)=0 x=−1⇒f(−1)=3+2=5 x=1⇒f(1)=3−2=1 Absolute minimum =(0,0) Absolute maximum =(−1,5)

Answered: "Find the absolute extrema of the function on..."

compagnia04

Answered question

2021-12-25

Find the absolute extrema of the function on the closed interval.

y = 3 x^{\frac{2}{3}} - 2 x, [- 1, 1]

minimum - ?
maximum - ?

Answer & Explanation

John Koga

Beginner2021-12-26Added 33 answers

y = 3 x^{\frac{2}{3}} - 2 x, [- 1, 1]

Now

\frac{d y}{d x} = 2 x^{\frac{1}{3}} - 2

= \frac{2}{x^{\frac{1}{3}}} - 2

= \frac{2 [1 - x^{\frac{1}{3}}]}{x^{\frac{1}{3}}}

x = 0 \Rightarrow f (0) = 0

x = - 1 \Rightarrow f (- 1) = 3 + 2 = 5

x = 1 \Rightarrow f (1) = 3 - 2 = 1

Absolute minimum

= (0, 0)

Absolute maximum

= (- 1, 5)

Joseph Fair

Beginner2021-12-27Added 34 answers

How do you go from setting the derivative of y to a critical point of x=0. Please show those steps. I am getting 1. Thanks

user_27qwe

Skilled2021-12-30Added 375 answers

I was struggling with this too, then after some toil remembered that critical numbers also include where the slope is undefined. The x=1 we kept getting is valid. Though, It is doubled up for evaluation by also being the endpoint on our closed interval. x=0 is where the solution is undefined, which only becomes apparent after combining the fraction!

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