Show that the function #f(x)=4x^2-5x# has a zero between #1# and #2#.

2 Answers

Answer:

As the value of function #f(x)# changes from negative to positive and as is continuous, it has a zero between #a# and #b#.

Explanation:

The function is #f(x)=4x^2-5x#.

Its derivative being #8x-5#, #f(x)# is a continuous function.

Further, at #a# i.e. #x=1#, the value of function is #4*1^2-5*1# or is #-1#.

At #b#, i.e. #x=2#, the value of function is #4*2^2-5*2# or is #6#.

It is obvious that between #a# and #b#, value of function changes from negative to positive and as it is continuous, it has a zero between #a# and #b#.

We may apply the Bolzano Theorem

Bolzano Theorem (BT)

Let, for two real a and b, a < b, a function f be continuous on a closed interval [a, b] such that f(a) and f(b) are of opposite signs. Then there exists a number #x_o#,in #[a, b]# with #f(x_o)=0#.

Hence we have that

#f(1)=4*1^2-5*1=-1#

#f(2)=4*2^2-5*2=6#

Because #f(1)*f(2)<0# then from Bolzano Theorem we have that there is a value #x_o# in #[1,2]# for which #f(x_o)=0#