Answer:
Step-by-step explanation:
A derivative is a limit and is expressed as a quotient where the denominator approaches zero. We input a function into this formula and come out with a slope formula. This slope finds the instantaneous rate of change, the instantaneous velocity, of the function which is, of course, the first derivative of the function.
If we use the limiting formula for a derivative for the function
[tex]f(x)=2x^2+3x+1[/tex] it fills in like this:
[tex]\frac{2(x+h)^2+3(x+h)+1-(2x^2+3x+1)}{h}[/tex] as h approaches 0. Simplifying a bit gives us:
[tex]\frac{2(x^2+2xh+h^2)+3x+3h+1-2x^2-3x-1}{h}[/tex] as h approaches 0.
The following terms will cancel each other out:
[tex]2x^2,3x,1[/tex], leaving us with:
(4xh + 2h^2 + 3h)/h as h approaches 0.
If we factor out an h on the top:
[tex]\frac{h(4x+2h+3)}{h}[/tex] as h approaches 0.
Canceling out the h's, we get that the derivative, the slope formula, is
f'(x) = 4x + 2h + 3 as h approaches 0. If h is approaching 0, then the term 2h is equal to 0, so the derivative is
f'(x) = 4x + 3.
Following the rules for the derivative, once you know them, it is easy to see the derivative is the same: 4x + 3
For part 2. the position function is
[tex]s(t)=-3+4t^2[/tex]
The instantaneous velocity is found by finding the derivative of this position function. The derivative is:
s'(t) = 8t
If we evaluate this velocity equation at a t value of 5, we get that the instantaneous velocity at 5 seconds is 40
For part 3. a function that would have that derivative could be:
[tex]f(x)=x^2+x+3[/tex]
