Given f(x) = -frac{x^3}{3} + x^2 sin(1.5a) - | vedclass

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(D) The domain of $\sin^{-1}(u)$ is $[-1, 1]$. Here,$u = a^2 - 8a + 17 = (a-4)^2 + 1$. For this to be defined,$(a-4)^2 + 1 \le 1$,which implies $(a-4)

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$f'( \sin(8) ) < 0$

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(D) The domain of $\sin^{-1}(u)$ is $[-1, 1]$. Here,$u = a^2 - 8a + 17 = (a-4)^2 + 1$. For this to be defined,$(a-4)^2 + 1 \le 1$,which implies $(a-4)^2 \le 0$. Since a square is always non-negative,this is only possible if $a = 4$.Substituting $a = 4$ into $f(x)$:$f(x) = -\frac{x^3}{3} + x^2 \sin(6) - x \sin(4) \sin(8) - 5 \sin^{-1}(1)$.Now,differentiate $f(x)$ with respect to $x$:$f'(x) = -x^2 + 2x \sin(6) - \sin(4) \sin(8)$.Evaluate $f'( \sin(8) )$:$f'( \sin(8) ) = -\sin^2(8) + 2 \sin(8) \sin(6) - \sin(4) \sin(8)$.$f'( \sin(8) ) = \sin(8) [ -\sin(8) + 2 \sin(6) - \sin(4) ]$.Using the identity $\sin(A) + \sin(B) = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:$f'( \sin(8) ) = -\sin(8) [ \sin(8) + \sin(4) - 2 \sin(6) ]$.$f'( \sin(8) ) = -\sin(8) [ 2 \sin(6) \cos(2) - 2 \sin(6) ]$.$f'( \sin(8) ) = -2 \sin(8) \sin(6) [ \cos(2) - 1 ] = 2 \sin(8) \sin(6) [ 1 - \cos(2) ]$.Since $\sin(8) \approx 0.989$ (positive),$\sin(6) \approx -0.279$ (negative),and $1 - \cos(2) > 0$,the product is negative.Thus,$f'( \sin(8) ) < 0$.

Given $f(x) = -\frac{x^3}{3} + x^2 \sin(1.5a) - x \sin(a) \sin(2a) - 5 \sin^{-1}(a^2 - 8a + 17)$,then:

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