9. -/5 points details my notes scalc9 2.9.027. consider the following. f(x)=x^4 - x + 5 compare the values…

Explanation:

Step1: Find the derivative of $f(x)$

$f(x)=x^{4}-x + 5$, so $f^\prime(x)=4x^{3}-1$.

Step2: Calculate $\Delta y$ when $x$ changes from $1$ to $1.07$

$\Delta y=f(1.07)-f(1)=(1.07^{4}-1.07 + 5)-(1^{4}-1 + 5)=1.07^{4}-1.07+5 - 5=1.07^{4}-1.07\approx1.3108 - 1.07=0.2408$.

Step3: Calculate $dy$ when $x$ changes from $1$ to $1.07$

$dx=\Delta x=1.07 - 1=0.07$, $f^\prime(1)=4\times1^{3}-1 = 3$, $dy=f^\prime(1)dx=3\times0.07 = 0.2100$.

Step4: Calculate $\Delta y$ when $x$ changes from $1$ to $1.01$

$\Delta y=f(1.01)-f(1)=(1.01^{4}-1.01 + 5)-(1^{4}-1 + 5)=1.01^{4}-1.01\approx1.0406 - 1.01=0.0306$.

Step5: Calculate $dy$ when $x$ changes from $1$ to $1.01$

$dx=\Delta x=1.01 - 1=0.01$, $f^\prime(1)=3$, $dy=f^\prime(1)dx=3\times0.01 = 0.0300$.

Answer:

When $x$ changes from $1$ to $1.07$: $\Delta y = 0.2408$ $dy = 0.2100$ When $x$ changes from $1$ to $1.01$: $\Delta y = 0.0306$ $dy = 0.0300$ Yes