# How can this derivative be solved?

#### lasvegas666

##### New member

They want you to find

[imath]\displaystyle \frac{df}{dx} = \frac{\Delta f}{\Delta x}[/imath]

They want you to find

[imath]\displaystyle \frac{df}{dx} = \frac{\Delta f}{\Delta x}[/imath]
How do I use that formula in the question?

How do I use that formula in the question?
By figuring out function [imath]\frac{df}{dx}[/imath] and the values of [imath]\Delta f[/imath] and [imath]\Delta x[/imath].

By figuring out function [imath]\frac{df}{dx}[/imath] and the values of [imath]\Delta f[/imath] and [imath]\Delta x[/imath].
I'm genuinely confused.

How do I solve this problem with df/dx?

I'm genuinely confused.

How do I solve this problem with df/dx?
Think about it. I am sure that you will figure it out.

Think about it. I am sure that you will figure it out.
...so it's 2x - 7?

It is. The notation [imath] dy/dx [/imath] reminds of the definition of the derivative, the slope of secants [imath] \Delta y / \Delta x [/imath] that converge to the slope of a tangent if [imath] \Delta x \to 0. [/imath] In formulas, given [imath] y=x^2-7x+ 3,[/imath] it is

$\begin{array}{lll} y'&=\dfrac{dy}{dx}=\lim_{\Delta x \to 0}\dfrac{\Delta y}{\Delta x}\\[12pt]&=\lim_{\Delta x \to 0}\dfrac{y(x+\Delta x)-y(x)}{\Delta x}\\[12pt] &=\lim_{\Delta x \to 0}\dfrac{(x+\Delta x)^2-7(x+\Delta x)+3 - (x^2-7x+3)}{\Delta x}\\[12pt] &=\lim_{\Delta x \to 0}\dfrac{x^2+2\cdot x\cdot \Delta x +(\Delta x)^2-7x-7\Delta x +3-x^2+7x-3}{\Delta x}\\[12pt] &=\lim_{\Delta x \to 0}(2\cdot x +\Delta x -7)=2x-7 \end{array}$

It is. The notation [imath] dy/dx [/imath] reminds of the definition of the derivative, the slope of secants [imath] \Delta y / \Delta x [/imath] that converge to the slope of a tangent if [imath] \Delta x \to 0. [/imath] In formulas, given [imath] y=x^2-7x+ 3,[/imath] it is

$\begin{array}{lll} y'&=\dfrac{dy}{dx}=\lim_{\Delta x \to 0}\dfrac{\Delta y}{\Delta x}\\[12pt]&=\lim_{\Delta x \to 0}\dfrac{y(x+\Delta x)-y(x)}{\Delta x}\\[12pt] &=\lim_{\Delta x \to 0}\dfrac{(x+\Delta x)^2-7(x+\Delta x)+3 - (x^2-7x+3)}{\Delta x}\\[12pt] &=\lim_{\Delta x \to 0}\dfrac{x^2+2\cdot x\cdot \Delta x +(\Delta x)^2-7x-7\Delta x +3-x^2+7x-3}{\Delta x}\\[12pt] &=\lim_{\Delta x \to 0}(2\cdot x +\Delta x -7)=2x-7 \end{array}$
Alright.

How do I factor the coordinates [0, 4] into this problem?

Okie Dokie

Thank you for not answering my question.

...so it's 2x - 7?
Looks like [imath]\frac{df}{dx}[/imath] If that's what you mean than it's correct.

I'm genuinely confused.

How do I solve this problem with df/dx?
I am guessing that @mario99 means [imath]\Delta f = f(x_1) - f(x_0)[/imath] and [imath]\Delta x = x_1 - x_0[/imath]. Does this make sense to you?

I am guessing that @mario99 means [imath]\Delta f = f(x_1) - f(x_0)[/imath] and [imath]\Delta x = x_1 - x_0[/imath]. Does this make sense to you?
I got -4 as my answer.

Yep.

How do I arrive at my final answer?
I think the average (integral over divided by distance) rate of change (first derivative) is given as
$\dfrac{1}{4-0}\int_0^4 y'(t)\,dt = \dfrac{1}{4}\left[y(t)\right]_0^4=\dfrac{4^2-7\cdot 4+3-3}{4}=-3$and the answer to when the instantaneous rate (derivative at a certain point) equals [imath] -3 [/imath] is therefore
$y'(x)=2x-7\stackrel{!}{=}-3\Longrightarrow x=2.$

They found for you the average rate of change over the interval from 0 to 4, $\frac{f(4)-f(0)}{4-0}=\frac{(4^2-7(4)+3)-(0^2-7(0)+3)}{4}=\frac{(-9)-(3)}{4}=\frac{-12}{4}=-3$
All you are asked to do here is to find a value of x between 0 and 4 for which the instantaneous rate of change, f'(x), is equal to -3, which the mean value theorem says must exist.

You've stated that your answer is x = -4. First, that is not between 0 and 4; second, the derivative you correctly stated, f'(x) = 2x - 7, is not equal to -3 at x = -4.