91Ó°ÊÓ

Evaluate the function \(f(x)\) when \(x=0\) in both left and right sides. For \(x \leq 0\), we have: $$f(x)=\sin \left(x^{2}-3 x\right)$$ Evaluate at \(x=0\): $$f(0) = \sin \left(0^{2} - 3(0)\right) = \sin(0) = 0$$ For \(x > 0\), we have: $$f(x)=6 x+5 x^{2}$$ Evaluate at \(x=0\): $$f(0) = 6(0) + 5(0)^{2} = 0$$ Since \(f(0)\) is the same when approaching from both left and right sides, the function is continuous at \(x = 0\). Now we will find if there is a local maximum or minimum at this point.

To determine if we have a local maximum or minimum at \(x=0\), we need to analyze the derivative of \(f(x)\) at this point. For \(x \leq 0\): $$f(x)=\sin \left(x^{2}-3 x\right)$$ Differentiate with respect to \(x\): $$f'(x)=\cos \left(x^{2}-3 x\right)(2x-3)$$ Evaluate at \(x=0\): $$f'(0) = \cos \left(0\right)(-3) = 1\cdot(-3) = -3$$ For \(x > 0\): $$f(x)=6 x+5 x^{2}$$ Differentiate with respect to \(x\): $$f'(x)=6+10 x$$ Evaluate at \(x=0\): $$f'(0) = 6+10(0) = 6$$ As we can see, the left-hand derivative \(f'(0^{-})=-3\) is negative and the right-hand derivative \(f'(0^{+})=6\) is positive. Since the function is continuous at \(x=0\), and we have a change in the sign of the derivative, the function \(f(x)\) has a local minimum at \(x=0\). So the correct answer is: (B) has a local minimum