How can this derivative be solved?

They want you to find

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

[math]\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}[/math]
 
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

[math]\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}[/math]
Alright.

How do I factor the coordinates [0, 4] into this problem?
 
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
[math] \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 [/math]and the answer to when the instantaneous rate (derivative at a certain point) equals [imath] -3 [/imath] is therefore
[math] y'(x)=2x-7\stackrel{!}{=}-3\Longrightarrow x=2. [/math]
 
They found for you the average rate of change over the interval from 0 to 4, [math]\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[/math]
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.
 
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