Math Help - Derivative help

1. Derivative help

Let
f(x)=2*x^2+5*x - 3
and let x0= 5.

A. The average rate of change of f between x= 5 and x= 5.14 equals ____

B. The average rate of change of f between x= 5 and x= 5.014 equals ___

C. The average rate of change of f between x= 5 and x= 5.0014 equals ____

D. The instantaneous rate of change of f at x= 5 equals ____

2. Originally Posted by lilwayne

Let
f(x)=2*x^2+5*x - 3
and let x0= 5.

A. The average rate of change of f between x= 5 and x= 5.14 equals ____

B. The average rate of change of f between x= 5 and x= 5.014 equals ___

C. The average rate of change of f between x= 5 and x= 5.0014 equals ____

D. The instantaneous rate of change of f at x= 5 equals ____
You really should post your efforts, thoughts or where you are stuck...

Hint: the average rate of change is the slope of the straight line joining the two points,

hence it is $\displaystyle\frac{f(x_2)-f(x_1)}{x_2-x_1}$

The instantaneous rate of change is given by the slope of the tangent at a particular f(x).
The derivative is the slope of the tangent evaluated at the given x.

3. that's what i don't get, what do i use the formula for, like the f(x2) comes from where? If i figure out how to do the first one, i could probably do the rest by my self.

4. Originally Posted by lilwayne

Let
f(x)=2*x^2+5*x - 3
and let x0= 5.

A. The average rate of change of f between x= 5 and x= 5.14 equals ____

B. The average rate of change of f between x= 5 and x= 5.014 equals ___

C. The average rate of change of f between x= 5 and x= 5.0014 equals ____

D. The instantaneous rate of change of f at x= 5 equals ____
Here's how it works...

(A)

$x_2>x_1\Rightarrow\ x_1=5,\;\;\;x_2=5.14$

$f\left(x_1\right)$ is the result you get when you place the value $x_1$ into $f(x)$

in other words, just replace $x$ with the value $x_1$ and calculate the result.

Do the same for $x_2$

Then you can calculate the average rate of change.
It's the same as calculating the slope of a line between two given points

$\displaystyle\ slope=\frac{y_2-y_1}{x_2-x_1}$

Hence, the solution for (A) is

$\displaystyle\frac{\left[2(5.14)^2+5(5.14)-3\right]-\left[2(5)^2+5(5)-3\right]}{5.14-5}$

(B) and (C) are evaluated in the same way.

How are you a calculating derivatives for part (D) ?

5. i calculated 46.71 and punched that in, but it said incorrect :S

6. oops calculated wrong, got it right now lol thanks