# 009B Sample Midterm 3, Problem 3 Detailed Solution

Find a curve  ${\displaystyle y=f(x)}$  with the following properties:

(i)   ${\displaystyle f''(x)=6x}$

(ii)   Its graph passes through the point  ${\displaystyle (0,1)}$  and has a horizontal tangent there.

Background Information:

1. If the graph of  ${\displaystyle f(x)}$  passes through the point  ${\displaystyle (a,b),}$  then  ${\displaystyle f(a)=b.}$

2. If  ${\displaystyle f(x)}$  has a horizontal tangent at the point  ${\displaystyle (a,b),}$  then  ${\displaystyle f'(a)=0.}$

Solution:

Step 1:
Since  ${\displaystyle f(x)}$  passes through the point  ${\displaystyle (0,1),}$
${\displaystyle f(0)=1.}$
Since  ${\displaystyle f(x)}$  has a horizontal tangent at  ${\displaystyle (0,1),}$
${\displaystyle f'(0)=0.}$
Step 2:
Now, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\int f''(x)~dx}\\&&\\&=&\displaystyle {\int 6x~dx}\\&&\\&=&\displaystyle {3x^{2}+C.}\\\end{array}}}$

Since  ${\displaystyle f'(0)=0,}$  we have

${\displaystyle {\begin{array}{rcl}\displaystyle {0}&=&\displaystyle {f'(0)}\\&&\\&=&\displaystyle {3(0)^{2}+C}\\&&\\&=&\displaystyle {C.}\\\end{array}}}$

Hence,
${\displaystyle f'(x)=3x^{2}.}$
Step 3:
We have

${\displaystyle {\begin{array}{rcl}\displaystyle {f(x)}&=&\displaystyle {\int f'(x)~dx}\\&&\\&=&\displaystyle {\int 3x^{2}~dx}\\&&\\&=&\displaystyle {x^{3}+D.}\\\end{array}}}$

Since  ${\displaystyle f(0)=1,}$  we have

${\displaystyle {\begin{array}{rcl}\displaystyle {1}&=&\displaystyle {f(0)}\\&&\\&=&\displaystyle {(0)^{3}+D}\\&&\\&=&\displaystyle {D.}\\\end{array}}}$

Hence,
${\displaystyle f(x)=x^{3}+1.}$

${\displaystyle f(x)=x^{3}+1}$