a)
"F(x)=8-3x+x^2\\\\\nF'(x)=-3+2x\\\\\n\\text{At critical point, derivative is 0}\\\\\n-3+2x=0\\\\\n2x=3\\\\\nx=1.5\\\\\nF(x)=8-3(1.5)+(1.5)^2=5.75\\\\\n\\text{Critical point is (1.5,5.75)}"
b)
"F(x)=x^4-18x^2+9\\\\\nF'(x)=4x^3-36x\\\\\n4x^3-36x=0\\\\\nx=0,3\\\\\n\\text{at x=0, F(x)=9}\\\\\n\\text{at x=3, F(x)=-72}\\\\\n\\text{Critical points are (0,9) and (3,-72)}"
c)
"F(x)=x^3-5x^2-8x+3\\\\\nF'(x)=3x^2-10x-8\\\\\n3x^2-10x-8=0\\\\\nx=4,\\frac{-2}{3}\\\\\n\\text{At x=4, F(x)=-2}\\\\\n\\text{At x=}\\frac{-2}{3}\\text{ F(x)=}\\frac{157}{27}\\\\\n\\text{Critical points are (4,-2) and } (\\frac{-2}3,\\frac{157}{27})\\\\\nd)\\\\\nF(x)=\\dfrac{x^2}{x^2+1}\\\\\n\\text{has no critical point}\\\\\ne)F(x)=\\dfrac{x^2}{x^2-1}\\\\\n\\text{has no critical point}\\\\\nf) F(x)=(x^2-1)^3\\\\\nF'(x)=2x(x^2-1)^2\\\\\n2x(x^2-1)^2=0\\\\\nx=0,-1,1\\\\\n\\text{critical points are (0,-1),(-1,0) and (1,0)}"
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