Question

Determine the absolute minimum value of $f(x) = x\sqrt{1 - x^2} - 3$ on $[-1, 1]$. 1. absolute min. value $= -\frac{5}{2}$ 2. absolute min. value $= -4$ 3. absolute min. value $= -\frac{7}{2}$ 4. absolute min. value $= -2$ 5. absolute min. value $= -\frac{9}{2}$ 6. absolute min. value $= -3$

          Determine the absolute minimum value of
$f(x) = x\sqrt{1 - x^2} - 3$
on $[-1, 1]$.
1. absolute min. value $= -\frac{5}{2}$
2. absolute min. value $= -4$
3. absolute min. value $= -\frac{7}{2}$
4. absolute min. value $= -2$
5. absolute min. value $= -\frac{9}{2}$
6. absolute min. value $= -3$

Determine the absolute minimum value of
f(x) = x√(1 - x^2) - 3
on [-1, 1].
1. absolute min. value = -(5)/(2)
2. absolute min. value = -4
3. absolute min. value = -(7)/(2)
4. absolute min. value = -2
5. absolute min. value = -(9)/(2)
6. absolute min. value = -3

Added by Briana M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Linda Hand

03/27/2022

Step 1

f(x) = x√(1-x^2) - 3 f'(x) = √(1-x^2) - x^2 / √(1-x^2) Show more…

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Determine the absolute minimum value of f(x)=xsqrt(1-x^(2))-3 on -1,1. absolute min. value =-(5)/(2) absolute min. value =-4 absolute min. value =-(7)/(2) absolute min. value =-2 absolute min. value =-(9)/(2) absolute min. value =-3 Determine the absolute minimum value of f(x) = xV1-x2-3 on [-1, 1]. 5 2 1. absolute min. value = 2. absolute min. value = -4 7 2 3. absolute min. value = 4. absolute min. value = -2 9 2 5. absolute min. value = 6. absolute min. value = -3