Let f(x) = #x^3+5x^2+17x-10#. The equation f(x) = 0 has only one real root. So how do you sow the root lies between 0 and 2?

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Answer 1 · 2017-09-26T14:09:04

See the explanation below

The function #f(x)# is continuous on the interval #[a,b]# such that #f(a)<0# and #f(b)>0#, then there exists #c in [a,b]# such that #f(c)=0#

Here,

#f(x)=x^3+5x^2+17x-10#

This is a polynomial function, continuous on the interval #x in [0,2]#

#f(0)=-10#, #=>#, #f(0)<0#

#f(2)=2^3+5*2^2+17*2-10=8+20+34-10=52#, #f(2)>0#

Therefore,

#EE c in [0,2]# such that #f(c)=0#

You can calculate the exact root by iteration