Answer: 2.25
Step-by-step explanation:
First, we will find the zeros of the function.
Given:
\(\displaystyle H(x)=4x^3-5x^2-23x+6\)
Set equal to zero:
\(\displaystyle 0=4x^3-5x^2-23x+6\)
Factor:
\(\displaystyle 0=(x-3)(4x^2+7x-2)\)
\(\displaystyle 0=(x-3)(x+2)(4x-1)\)
Zero product property:
\(x=3,-2,\dfrac{1}{4}\)
Next, we will find which two zeros, also known as solutions, are the closest together.
| 3 - -2 | = 5
| -2 - 1/4 | = 9/4 = 2.25
| 1/4 - 3 | = 13/4 = 3.25
\(\displaystyle H(\frac{1}{4})=4(\frac{1}{4})^3-5(\frac{1}{4})^2-23(\frac{1}{4})+6=0\)
Finally, we will find the distance between the two closest together. These zeros are -2 and one-fourth. We can use the distance formula for this. For the zeros, both of the y-values equal zero.
\(\displaystyle d=\sqrt{(x_{2}-x_{1})^2 +(y_{2}-y_{1})^2}\)
\(\displaystyle d=\sqrt{(-2-\frac{1}{4} )^2 +(0-0)^2}\)
\(\displaystyle d=\sqrt{(-2-\frac{1}{4})^2\)
\(\displaystyle d=\sqrt{(-2.25)^2\)
\(\displaystyle d=\sqrt{5.0625}\)
\(\displaystyle d=2.25\)
