What values(s) of the constant b make f(x)=x^{3}−bx for 0leq xleq 2 have: a.) an absolute min at x=1? Explain. b.) an absolute max at x=2? Explain.

Bergen

Bergen

Answered question

2021-02-13

What values(s) of the constant b make f(x)=x3bx for 0x2 have:
a.) an absolute min at x=1? Explain.
b.) an absolute max at x=2? Explain.

Answer & Explanation

krolaniaN

krolaniaN

Skilled2021-02-14Added 86 answers

(a). Need to find b so that f(x)=x3bxf(x)=(x3)bx has minimum value at x=1.
Here, f(x)=(3x2)b. Now f'(1) = 0 gives us3b=0b=3
Here we have
f(x)=6xf(1)=6,
therefore, ff has an absolute min at x=1.
(b). We have f=3(x2)b, so f is increasing whenever b is non-negative, In that case, f is increasing in 0x2 and attains its absolute maximum at x=2.

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