Question

consider the following function.

$f(x)=48x$

find $f(x)$. (use $c$ for the constant of the first antiderivative.)

$f(x)=$

find $f$. (use $d$ for the constant of the second antiderivative.)

$f(x)=$

Explanation:

Step1: Find \(f'(x)\)

Recall the power rule for integration \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)).
Given \(f''(x)=48x\), then \(f'(x)=\int f''(x)dx=\int48x dx\).
Using the constant - multiple rule \(\int k\cdot g(x)dx=k\int g(x)dx\) (\(k = 48\), \(g(x)=x\)), we have \(f'(x)=48\int xdx\).
Applying the power rule with \(n = 1\), \(48\int xdx=48\times\frac{x^{1+1}}{1 + 1}+C\).
Simplify the expression: \(f'(x)=24x^{2}+C\).

Step2: Find \(f(x)\)

Since \(f'(x)=24x^{2}+C\), then \(f(x)=\int f'(x)dx=\int(24x^{2}+C)dx\).
Using the sum rule \(\int(a(x)+b(x))dx=\int a(x)dx+\int b(x)dx\), we get \(f(x)=\int24x^{2}dx+\int Cdx\).
For \(\int24x^{2}dx\), using the constant - multiple rule and power rule (\(n = 2\)), \(24\int x^{2}dx=24\times\frac{x^{2+1}}{2+1}\).
For \(\int Cdx\), using the rule \(\int kdx=kx + D\) (\(k = C\)), \(\int Cdx = Cx+D\).
So \(f(x)=8x^{3}+Cx + D\).

Answer:

\(f'(x)=24x^{2}+C\)
\(f(x)=8x^{3}+Cx + D\)