Find $f ( x )$ if $f ( 0 ) = 0$ and the tangent line at $( x , f ( x ) )$ has the slope $x \left( x ^ { 2 } + 2 \right) ^ { 3 }$ .
A) $\frac { x ^ { 2 } + 2 } { 8 } + C$
B) $\frac { x ^ { 2 } + 2 } { 8 }$
C) $\frac { \left( x ^ { 2 } + 2 \right) ^ { 4 } } { 4 } + 4$
D) $\frac { \left( x ^ { 2 } + 2 \right) ^ { 4 } } { 8 } - 2$
E) $\frac { \left( x ^ { 2 } + 2 \right) ^ { 4 } } { 4 } - 4$

Question

Find $f ( x )$ if $f ( 0 ) = 0$ and the tangent line at $( x , f ( x ) )$ has the slope $x \left( x ^ { 2 } + 2 \right) ^ { 3 }$ . 
A) $\frac { x ^ { 2 } + 2 } { 8 } + C$
B)  $\frac { x ^ { 2 } + 2 } { 8 }$
C)  $\frac { \left( x ^ { 2 } + 2 \right) ^ { 4 } } { 4 } + 4$
D)  $\frac { \left( x ^ { 2 } + 2 \right) ^ { 4 } } { 8 } - 2$
E)   $\frac { \left( x ^ { 2 } + 2 \right) ^ { 4 } } { 4 } - 4$

Quizplus · Accepted Answer

The slope of the tangent line at $(x, f(x))$ is given by the derivative of $f(x)$ with respect to $x$, which is $f'(x) = x(x^2 + 2)^3$. To find $f(x)$, we integrate $f'(x)$ with respect to $x$. The integral of $x(x^2 + 2)^3$ with respect to $x$ is $\frac{(x^2 + 2)^4}{8}$ plus a constant of integration. Since $f(0) = 0$, we can solve for the constant by substituting $x = 0$ into $f(x) = \frac{(0^2 + 2)^4}{8} + C$, which simplifies to $0 = \frac{16}{8} + C$, or $C = -2$. Therefore, $f(x) = \frac{(x^2 + 2)^4}{8} - 2$.

Find f(x)f ( x )f(x) If f(0)=0f ( 0 ) = 0f(0)=0

Find $f ( x )$ If $f ( 0 ) = 0$