Question

for y = f(x)=x^{4}-9x^{3}+6, find dy and δy, given x = 3 and δx=-0.2.
dy = (type an integer or a decimal.)

Explanation:

Step1: Find the derivative of y

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, for $y = x^{4}-9x^{3}+6$, we have $y^\prime=f^\prime(x)=4x^{3}-27x^{2}$.

Step2: Calculate dy

The differential $dy = f^\prime(x)\Delta x$. Substitute $x = 3$ and $\Delta x=-0.2$ into $f^\prime(x)$. First, find $f^\prime(3)$: $f^\prime(3)=4\times3^{3}-27\times3^{2}=4\times27 - 27\times9=108 - 243=-135$. Then $dy=f^\prime(3)\Delta x=(-135)\times(-0.2)=27$.

Step3: Calculate $\Delta y$

$\Delta y=f(x + \Delta x)-f(x)$. Here, $x = 3$ and $\Delta x=-0.2$, so $x+\Delta x=3+( - 0.2)=2.8$. $f(3)=3^{4}-9\times3^{3}+6=81-243 + 6=-156$. $f(2.8)=2.8^{4}-9\times2.8^{3}+6=61.4656-9\times21.952+6=61.4656 - 197.568+6=-130.1024$. Then $\Delta y=f(2.8)-f(3)=-130.1024-(-156)=25.8976$.

Answer:

$dy = 27$
$\Delta y=25.8976$