# Use the difference quotient to calculate the average rate of change across the following intervals. Difference quotient of d(t): 3t2 + 5th – 2 The interval 2 to 3: The interval 2 to 2.5: The interval 2 to 2.1:

Use the difference quotient to calculate the average rate of change across the following intervals.
Difference quotient of $d\left(t\right):3{t}^{2}+5th–2$
The interval 2 to 3:
The interval 2 to 2.5:
The interval 2 to 2.1:
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Leroy Cunningham
(i) The difference quotient for the interval 2 to 3 is 20.
(ii) The difference quotient for the interval 2 to 2.5 is 18.5.
(iii) The difference quotient for the interval 2 to 2.1 is 17.3.
Given a Function d(t) that is Continuous at Interval [a,b], the Difference Quotient associated to the Interval is:
${q}_{ab}=\frac{d\left(b\right)-d\left(a\right)}{b-a}\left(1\right)$
Where:
d(a)- Function evaluated at lower bound.
d(b)- Function evaluated at upper bound.
In this question, the function is represented by $d\left(t\right)=3\cdot {t}^{2}+5\cdot t-2$
(i) If we know that and , then the difference quotient is:
${q}_{ab}=\frac{3\cdot \left(3{\right)}^{2}+5\cdot \left(3\right)-2-\left[3\cdot \left(2{\right)}^{2}+5\cdot \left(2\right)-2\right]}{3-2}\phantom{\rule{0ex}{0ex}}{q}_{ab}=20$
The difference quotient for the interval 2 to 3 is 20.
(ii) If we know that a=2 and b=2.5, then the difference quotient is:
${q}_{ab}=\frac{3\cdot \left(2.5{\right)}^{2}+5\cdot \left(2.5\right)-2-\left[3\cdot \left(2{\right)}^{2}+5\cdot \left(2\right)-2\right]}{2.5-2}\phantom{\rule{0ex}{0ex}}{q}_{ab}=18.5$
The difference quotient for the interval 2 to 2.5 is 18.5.
(iii) If we know that a=2 and b=2.1, then the difference quotient is:
${q}_{ab}=\frac{3\cdot \left(2.1{\right)}^{2}+5\cdot \left(2.1\right)-2-\left[3\cdot \left(2{\right)}^{2}+5\cdot \left(2\right)-2\right]}{2.1-2}\phantom{\rule{0ex}{0ex}}{q}_{ab}=17.3$
The difference quotient for the interval 2 to 2.1 is 17.3.