Question

for ( y = f(x) = 6x^{3}, x = 5 ), and ( delta x = 0.02 ) find
a) ( delta y ) for the given ( x ) and ( delta x ) values,
b) ( dy = f(x)dx ),
c) ( dy ) for the given ( x ) and ( delta x ) values.
a) ( delta y=)
(round to four decimal places as needed.)
b) ( dy = f(x)dx = ) dx
c) ( dy=)
(round to two decimal places as needed.)

Explanation:

Step1: Calculate \(\Delta y\)

We know that \(\Delta y=f(x + \Delta x)-f(x)\).
Given \(f(x)=6x^{3}\), \(x = 5\), and \(\Delta x=0.02\).
First, find \(f(x+\Delta x)=6(5 + 0.02)^{3}=6\times(5.02)^{3}\).
Using the formula \((a + b)^{3}=a^{3}+3a^{2}b + 3ab^{2}+b^{3}\), where \(a = 5\) and \(b=0.02\), we have \((5.02)^{3}=5^{3}+3\times5^{2}\times0.02+3\times5\times(0.02)^{2}+(0.02)^{3}=125+1.5 + 0.006+0.000008 = 126.506008\).
Then \(f(x+\Delta x)=6\times126.506008 = 759.036048\).
And \(f(x)=6\times5^{3}=6\times125 = 750\).
So \(\Delta y=759.036048-750=9.036048\approx9.0360\).

Step2: Find the derivative \(f^{\prime}(x)\)

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), for \(y = f(x)=6x^{3}\), \(f^{\prime}(x)=\frac{d}{dx}(6x^{3})=6\times3x^{2}=18x^{2}\).
So \(dy=f^{\prime}(x)dx = 18x^{2}dx\).

Step3: Calculate \(dy\)

Substitute \(x = 5\) and \(dx=\Delta x = 0.02\) into \(dy = 18x^{2}dx\).
\(dy=18\times5^{2}\times0.02\).
First, \(5^{2}=25\), then \(18\times25\times0.02=(18\times25)\times0.02 = 450\times0.02=9.00\).

Answer:

a) \(\Delta y\approx9.0360\)
b) \(dy = 18x^{2}dx\)
c) \(dy = 9.00\)