(a). Need to find b so that \(f(x)=x3−bxf(x)=(x^{3})−bx\) has minimum value at x=1.

Here, \(f'(x) = (3x^{2})-b\). Now f'(1) = 0 gives us \(3-b=0 \Rightarrow b=3\)

Here we have

\(f''(x)=6x\Rightarrow f''(1)=6,\)

therefore, ff has an absolute min at x=1.

(b). We have \(f'=3(x^{2})-b\), so f is increasing whenever b is non-negative, In that case, f is increasing in \(0\leq x\leq2\) and attains its absolute maximum at x=2.

Here, \(f'(x) = (3x^{2})-b\). Now f'(1) = 0 gives us \(3-b=0 \Rightarrow b=3\)

Here we have

\(f''(x)=6x\Rightarrow f''(1)=6,\)

therefore, ff has an absolute min at x=1.

(b). We have \(f'=3(x^{2})-b\), so f is increasing whenever b is non-negative, In that case, f is increasing in \(0\leq x\leq2\) and attains its absolute maximum at x=2.