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| 1. |
Show that f(x)=x^3-6x^2+24x+4 has neither a maximum nor a minimum value. |
| Answer» Solution :`f(x)=x^3-6x^2+24+4=f(x)=3x^2-12x+24=3(x^2-4x+8)=3((x-2)^2+4)` as `f(x)ne`0 for all `x in R` the function has NEITHER a local MAXIMUM nor local minum. But has a ABSOLUTE minimum 12 at x =2. | |