How can this derivative be solved? f(x)=x^2-7x+3, average rate of change on [0,4] is -3

They want you to find

dfdx=ΔfΔx\displaystyle \frac{df}{dx} = \frac{\Delta f}{\Delta x}
 
It is. The notation dy/dx dy/dx reminds of the definition of the derivative, the slope of secants Δy/Δx \Delta y / \Delta x that converge to the slope of a tangent if Δx→0. \Delta x \to 0. In formulas, given y=x2−7x+3, y=x^2-7x+ 3, it is

y′=dydx=lim⁡Δx→0ΔyΔx=lim⁡Δx→0y(x+Δx)−y(x)Δx=lim⁡Δx→0(x+Δx)2−7(x+Δx)+3−(x2−7x+3)Δx=lim⁡Δx→0x2+2⋅x⋅Δx+(Δx)2−7x−7Δx+3−x2+7x−3Δx=lim⁡Δx→0(2⋅x+Δx−7)=2x−7\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}
 
fresh_42 said:
It is. The notation dy/dx dy/dx reminds of the definition of the derivative, the slope of secants Δy/Δx \Delta y / \Delta x that converge to the slope of a tangent if Δx→0. \Delta x \to 0. In formulas, given y=x2−7x+3, y=x^2-7x+ 3, it is

y′=dydx=lim⁡Δx→0ΔyΔx=lim⁡Δx→0y(x+Δx)−y(x)Δx=lim⁡Δx→0(x+Δx)2−7(x+Δx)+3−(x2−7x+3)Δx=lim⁡Δx→0x2+2⋅x⋅Δx+(Δx)2−7x−7Δx+3−x2+7x−3Δx=lim⁡Δx→0(2⋅x+Δx−7)=2x−7\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}
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Alright.

How do I factor the coordinates [0, 4] into this problem?
 
lasvegas666 said:
Yep.

How do I arrive at my final answer?
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I think the average (integral over divided by distance) rate of change (first derivative) is given as
14−0∫04y′(t) dt=14[y(t)]04=42−7⋅4+3−34=−3 \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 −3 -3 is therefore
y′(x)=2x−7=!−3⟹x=2. y'(x)=2x-7\stackrel{!}{=}-3\Longrightarrow x=2.
 
lasvegas666 said:
View attachment 37850
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They found for you the average rate of change over the interval from 0 to 4, f(4)−f(0)4−0=(42−7(4)+3)−(02−7(0)+3)4=(−9)−(3)4=−124=−3\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.
 
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