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

Feb 16, 2016

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

#### Explanation:

The function is $f \left(x\right) = 4 {x}^{2} - 5 x$.

Its derivative being $8 x - 5$, $f \left(x\right)$ is a continuous function.

Further, at $a$ i.e. $x = 1$, the value of function is $4 \cdot {1}^{2} - 5 \cdot 1$ or is $- 1$.

At $b$, i.e. $x = 2$, the value of function is $4 \cdot {2}^{2} - 5 \cdot 2$ 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 $\left[a , b\right]$ with $f \left({x}_{o}\right) = 0$.

Hence we have that

$f \left(1\right) = 4 \cdot {1}^{2} - 5 \cdot 1 = - 1$

$f \left(2\right) = 4 \cdot {2}^{2} - 5 \cdot 2 = 6$

Because $f \left(1\right) \cdot f \left(2\right) < 0$ then from Bolzano Theorem we have that there is a value ${x}_{o}$ in $\left[1 , 2\right]$ for which $f \left({x}_{o}\right) = 0$