a particle's motion is described by the equation d=t^2-8t+15 where d and t are measured in meters and seconds. show that the particle is at rest when t=4.
Again, if you know derivatives, then you can take the derivative of d(t) to get
$\displaystyle d(t) = 2t - 8$
... and then plug in t = 4 to see if the particle is indeed at rest.
Or, mimicking your previous post, find d(4) and d(4 + h), and plug into
$\displaystyle \lim_{h \to 0} \dfrac{d(4 + h) - d(4)}{h}$
and take the limit.
What function are you using? You said that $\displaystyle f(t)= t^2-8t+15$ so that f(4)= 16- 32+ 15= 1, not 7.
And f(4+h) is NOT $\displaystyle h^2+ 7$. Where did you get that?
Since $\displaystyle f(t)= t^2-8t+15$, $\displaystyle f(4+h)= (4+ h)^2- 8(4+ h)+ 15$
$\displaystyle f(4+h)= 16+ 8h+ h^2- 32- 8h+ 15= h^2+ 8h+ 1$
$\displaystyle \frac{f(4+h)- f(4)}{h}= \frac{h^2+ 8h+ 1- 1}{h}= \frac{h^2+ 8h}{h}$
and, for $\displaystyle h\ne 0$, that is h+ 8. Take the limit of that as h goes to 0.
No, you need to differentiate the position function to get velocity.
And it was not clear whether the OP could use "rules" of differentiation like that. Since this was posted in "Precalculus" it is likely that he knows only the "limit of the difference quotient" definition of derivative- or that he could find where the slope of the tangent line is 0 by completing the square as Captain Black suggested.we need to integrate velocity which comes out to be 2t-8
v=2t-8
v(4)=2(4)-8
=8-8
=0
hence proved
As mentioned elsewhere the the curve represented by the equation of displacement against time is a parabola and the particle is at rest when the tangent to the parabola is horizontal and this occurs at the vetex of the parabola.
The value of $\displaystyle t$ corresponding to the vertex is the mid point between the roots of the quadratic obtained by setting displacement to zero:
$\displaystyle t^2-8t+15=0$
Applying the quadratic formula to this quadratic give the roots as $\displaystyle t=4 \pm 1$
QED
Nope. Careful with your signs...
$\displaystyle \lim_{h \to 0} \dfrac{d(4 + h) - d(4)}{h}$
$\displaystyle = \lim_{h \to 0} \dfrac{h^2 - 1 - (-1)}{h}$
$\displaystyle = \lim_{h \to 0} \dfrac{h^2 - 1 + 1}{h}$
$\displaystyle = \lim_{h \to 0} \dfrac{h^2}{h}$
$\displaystyle = \lim_{h \to 0} h$
$\displaystyle = 0$