Question

3 slopes
find the instantaneous slope f(x) of the following functions. recall that in class we showed that, for f(x) = ∑_{n = 0}^{n} c_{n}x^{n}, f(x) = ∑_{n = 0}^{n} n c_{n}x^{n - 1}

1. f(x)=x^{3}+x^{2}+x + 1
2. f(x)=\frac{x^{4}}{4}-\frac{x^{2}}{2}+1
3. f(x)=x^{-2}-x^{-1}
4. f(x)=2x^{3}+3x^{-4}

for the following positions x(t), find the instantaneous velocity v(t). recall that the instantaneous velocity is the slope of x(t):

1. x(t)=x_{0}+v_{0}t+\frac{at^{2}}{2}
2. x(t)=\frac{at^{2}}{2}+\frac{bt^{3}}{3}
3. x(t)=v_{0}t+\frac{pt_{0}}{q}
4. x(t)=\frac{st^{4}}{4}+\frac{ct^{3}}{3}

Explanation:

Step1: Recall the power - rule of differentiation

The power - rule states that if $y = x^n$, then $y^\prime=nx^{n - 1}$.

Step2: Differentiate $f(x)=x^3+x^2+x + 1$

For $y=x^3$, $y^\prime = 3x^{2}$; for $y=x^2$, $y^\prime=2x$; for $y = x$, $y^\prime=1$; for $y = 1$, $y^\prime = 0$. So $f^\prime(x)=3x^{2}+2x + 1$.

Step3: Differentiate $f(x)=x^4/4 - x^2/2+1$

For $y=\frac{x^4}{4}$, $y^\prime=\frac{4x^{3}}{4}=x^{3}$; for $y=-\frac{x^2}{2}$, $y^\prime=-x$; for $y = 1$, $y^\prime = 0$. So $f^\prime(x)=x^{3}-x$.

Step4: Differentiate $f(x)=x^{-2}-x^{-1}$

For $y=x^{-2}$, $y^\prime=-2x^{-3}$; for $y=x^{-1}$, $y^\prime=-x^{-2}$. So $f^\prime(x)=-2x^{-3}+x^{-2}$.

Step5: Differentiate $f(x)=2x^{3}+3x^{-4}$

For $y = 2x^{3}$, $y^\prime=6x^{2}$; for $y = 3x^{-4}$, $y^\prime=-12x^{-5}$. So $f^\prime(x)=6x^{2}-12x^{-5}$.

1. $f^\prime(x)=3x^{2}+2x + 1$
2. $f^\prime(x)=x^{3}-x$
3. $f^\prime(x)=-2x^{-3}+x^{-2}$
4. $f^\prime(x)=6x^{2}-12x^{-5}$

Answer:

Step1: Recall the power - rule of differentiation

The power - rule states that if $y = x^n$, then $y^\prime=nx^{n - 1}$.

Step2: Differentiate $f(x)=x^3+x^2+x + 1$

For $y=x^3$, $y^\prime = 3x^{2}$; for $y=x^2$, $y^\prime=2x$; for $y = x$, $y^\prime=1$; for $y = 1$, $y^\prime = 0$. So $f^\prime(x)=3x^{2}+2x + 1$.

Step3: Differentiate $f(x)=x^4/4 - x^2/2+1$

For $y=\frac{x^4}{4}$, $y^\prime=\frac{4x^{3}}{4}=x^{3}$; for $y=-\frac{x^2}{2}$, $y^\prime=-x$; for $y = 1$, $y^\prime = 0$. So $f^\prime(x)=x^{3}-x$.

Step4: Differentiate $f(x)=x^{-2}-x^{-1}$

For $y=x^{-2}$, $y^\prime=-2x^{-3}$; for $y=x^{-1}$, $y^\prime=-x^{-2}$. So $f^\prime(x)=-2x^{-3}+x^{-2}$.

Step5: Differentiate $f(x)=2x^{3}+3x^{-4}$

For $y = 2x^{3}$, $y^\prime=6x^{2}$; for $y = 3x^{-4}$, $y^\prime=-12x^{-5}$. So $f^\prime(x)=6x^{2}-12x^{-5}$.

1. $f^\prime(x)=3x^{2}+2x + 1$
2. $f^\prime(x)=x^{3}-x$
3. $f^\prime(x)=-2x^{-3}+x^{-2}$
4. $f^\prime(x)=6x^{2}-12x^{-5}$