π(π₯) = π₯
3 β 12π₯, π₯ β [0, 4]
Ans:- "f(x)=x^3-12x" , "\ud835\udc65 \u2208 [0, 4]"
"\\Rightarrow" "f(x)=x(x^2-12)\\\\"
"\\Rightarrow f(x)=x(x+2\\sqrt{3})(x-2\\sqrt3)"
"\\Rightarrow f(x)=x(x+3.4641)(x-3.4641)"
We know that "x" lies in the interval between "\ud835\udc65 \u2208 [0, 4]".
Hence The values of "x" for which "f(x)" should be zero is "0, 3.4641" and
"f'(x)=3x^2-12"
"\\Rightarrow" "f'(0)=-12<0" so at "x=0" the slope of the f(x) is increasing
"\\Rightarrow f'(3.4641)=-5.5692<0" so at "x=3.4641" the slope of the "f(x)" is increasing
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