31–32. Velocity functions A projectile is fired vertically upward into the air, and its position (in feet) above the ground after t seconds is given by the function s(t).
b. Determine the instantaneous velocity of the projectile at t=1 and t = 2 seconds.
s(t)= −16t²+100t

Find the function The following limits represent the slope of a curve y = f(x) at the point (a,f(a)). Determine a possible function f and number a; then calculate the limit.

(lim h🠂0) (2+h)⁴-16 / h

The following table gives the distance f(t) fallen by a smoke jumper seconds after she opens her chute. <IMAGE>

a. Use the forward difference quotient with ℎ = 0.5 to estimate the velocity of the smoke jumper at t=2 seconds.

{Use of Tech} Approximating derivatives Assuming the limit exists, the definition of the derivative f′(a) = lim h→0 f(a + h) − f(a) / h implies that if ℎ is small, then an approximation to f′(a) is given by

f' (a) ≈ f(a+h) - f(a) / h. If ℎ > 0 , then this approximation is called a forward difference quotient; if ℎ < 0 , it is a backward difference quotient. As shown in the following exercises, these formulas are used to approximate f′ at a point when f is a complicated function or when f is represented by a set of data points. <IMAGE>

Let f (x) = √x.

a. Find the exact value of f' (4).

31–32. Velocity functions A projectile is fired vertically upward into the air, and its position (in feet) above the ground after t seconds is given by the function s(t).

a. For the following functions s(t), find the instantaneous velocity function v(t). (Recall that the velocity function v is the derivative of the position function s.)

s(t)= −16t²+100t

Calculator limits Use a calculator to approximate the following limits.

lim x🠂0 e^3x-1 / x

The speed of sound (in m/s) in dry air is approximated the function v(T) = 331 + 0.6T, where T is the air temperature (in degrees Celsius). Evaluate v' (T) and interpret its meaning.

The right-sided and left-sided derivatives of a function at a point $a$ are given by $f_{+}^{\prime }\left(a\right)=\underset{h\to 0^{+}}{\lim }\frac{f(a+h)-f\left(a\right)}{h}$ and $f_{-}^{\prime }\left(a\right)=\underset{h\to 0^{-}}{\lim }\frac{f(a+h)-f\left(a\right)}{h}$, respectively, provided these limits exist. The derivative $f^{\prime }\left(a\right)$ exists if and only if $f_{+}^{\prime }\left(a\right)=f_{-}^{\prime }\left(a\right)$.

Compute $f_{+}^{\prime }\left(a\right)$ and $f_{-}^{\prime }\left(a\right)$ at the given point $a$.

$f\left(x\right)=\mid x-2\mid$; $a=2$

The right-sided and left-sided derivatives of a function at a point $a$ are given by $f_{+}^{\prime }\left(a\right)=\underset{h\to 0^{+}}{\lim }\frac{f(a+h)-f\left(a\right)}{h}$ and $f_{-}^{\prime }\left(a\right)=\underset{h\to 0^{-}}{\lim }\frac{f(a+h)-f\left(a\right)}{h}$, respectively, provided these limits exist. The derivative $f^{\prime}\left(a\right)$ exists if and only if $f_{+}^{\prime}\left(a\right)=f_{-}^{\prime}\left(a\right)$.

Compute $f_{+}^{\prime}\left(a\right)$ and $f_{-}^{\prime}\left(a\right)$ at the given point $a$.

$f\left(x\right)=\{\begin{array}{c}4-x^2\text{ if }x\le 1\\ 2x+1\text{ if }x>1\end{array}$; $a=1$