lasvegas666
New member
- Joined
- Mar 1, 2024
- Messages
- 21
How do I use that formula in the question?They want you to find
[imath]\displaystyle \frac{df}{dx} = \frac{\Delta f}{\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].How do I use that formula in the question?
I'm genuinely confused.By figuring out function [imath]\frac{df}{dx}[/imath] and the values of [imath]\Delta f[/imath] and [imath]\Delta x[/imath].
Think about it. I am sure that you will figure it out.I'm genuinely confused.
How do I solve this problem with df/dx?
...so it's 2x - 7?Think about it. I am sure that you will figure it out.
What is this?...so it's 2x - 7?
How do I get the answer to this problem?What is this?
I asked you a question!How do I get the answer to this problem?
Alright.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]
Looks like [imath]\frac{df}{dx}[/imath] If that's what you mean than it's correct....so it's 2x - 7?
I got -4 as my answer.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?
Do you mean x = -4 ?I got -4 as my answer.
I think the average (integral over divided by distance) rate of change (first derivative) is given asYep.
How do I arrive at my final answer?
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]