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Let $f(x)=3 \cdot x^2+5 \cdot x-4$

The average rate of change of $f$ between $x = 1$ and $x = 1.17$ equals...?

The instantaneous rate of change of $f$ at $x = 1$ equals...?

How can I answer these two questions?

AtilioA
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jayson
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  • The average rate of change is $\frac{f(1.17)-f(1)}{1.17-1}$. Now calculate. For instantaneous rate of change, differentiate, evaluate the derivative at $t=1$. – André Nicolas Sep 23 '13 at 05:17
  • @AndréNicolas Not a 100% sure what you mean. Now calculate? Dont you have to find the derivative of the formula above? then plug in the numbers? Mind showing me how to do the first one? then I have a bit more of an idea. Much appreciated – jayson Sep 23 '13 at 05:31
  • The comment gave an explicit expression. You need to evaluate it, using the function you were given. Basically plugging in. We do not need the derivative for the average rate of change. For instantaneous rate of change, we do. That was the second part of the comment. – André Nicolas Sep 23 '13 at 05:40

1 Answers1

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$1.$ The average rate of change of $f(x)$ from $x=a$ to $x=b$ is $$\frac{f(b)-f(a)}{b-a}.$$ In our case $a=1$, $b=1.17$, and $f(x)=3x^2+5x-4$.

We have $f(1.17)=3(1.17)^2 +5(1.17)-4$. I think this is $5.9567$. Also, $f(1)=4$. Calculate.

$2.$ For the instantaneous rate of change at $x=1$, we need to compute $f'(1)$, the derivative of $f(x)$ at $x=1$.

The answers you get in ($1.$) and ($2.$) will not be equal, but will be fairly close to each other.

André Nicolas
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