Find $f ( x )$ if $f ( 1 ) = \frac { 11 } { 2 }$ and the tangent line at $( x , f ( x ) )$ has the slope $11 x e ^ { x ^ { 2 } - 1 }$ .
A) $\frac { 11 } { 2 } e ^ { x ^ { 2 } - 1 } + \frac { 11 } { 2 }$
B) $\frac { 1 } { 2 } e ^ { x ^ { 2 } - 1 } + C$
C) $\frac { 1 } { 2 } e ^ { x ^ { 2 } - 1 }$
D) $\frac { 11 } { 2 } e ^ { x ^ { 2 } - 1 }$
E) $11 e ^ { x ^ { 2 } - 1 }$

Question

Find $f ( x )$ if $f ( 1 ) = \frac { 11 } { 2 }$ and the tangent line at $( x , f ( x ) )$ has the slope $11 x e ^ { x ^ { 2 } - 1 }$ . 
A)  $\frac { 11 } { 2 } e ^ { x ^ { 2 } - 1 } + \frac { 11 } { 2 }$
B)  $\frac { 1 } { 2 } e ^ { x ^ { 2 } - 1 } + C$
C)  $\frac { 1 } { 2 } e ^ { x ^ { 2 } - 1 }$
D)  $\frac { 11 } { 2 } e ^ { x ^ { 2 } - 1 }$
E)   $11 e ^ { x ^ { 2 } - 1 }$

Quizplus · Accepted Answer

The derivative of $f(x)$ is given as $f'(x) = 11x e^{x^2 - 1}$. Integrating this, we get $f(x) = \frac{11}{2} e^{x^2 - 1} + C$. Using the condition $f(1) = \frac{11}{2}$, we find $C = 0$. Thus, $f(x) = \frac{11}{2} e^{x^2 - 1}$.

Find f(x)f ( x )f(x) If f(1)=112f ( 1 ) = \frac { 11 } { 2 }f(1)=211​

Find $f ( x )$ If $f ( 1 ) = \frac { 11 } { 2 }$