009B Sample Midterm 3, Problem 3 Detailed Solution

Find a curve  $y=f(x)$  with the following properties:

(i)   $f''(x)=6x$

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

---

ExpandBackground Information:
1. If the graph of  $f(x)$  passes through the point  $(a,b),$  then  $f(a)=b.$
2. If  $f(x)$  has a horizontal tangent at the point  $(a,b),$  then  $f'(a)=0.$

Solution:

ExpandStep 1:
Since  $f(x)$  passes through the point  $(0,1),$
$f(0)=1.$
Since  $f(x)$  has a horizontal tangent at  $(0,1),$
$f'(0)=0.$

ExpandStep 2:
Now, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\int f''(x)~dx}\\&&\\&=&\displaystyle {\int 6x~dx}\\&&\\&=&\displaystyle {3x^{2}+C.}\\\end{array}}$
Since  $f'(0)=0,$  we have
${\begin{array}{rcl}\displaystyle {0}&=&\displaystyle {f'(0)}\\&&\\&=&\displaystyle {3(0)^{2}+C}\\&&\\&=&\displaystyle {C.}\\\end{array}}$
Hence,
$f'(x)=3x^{2}.$

ExpandStep 3:
We have
${\begin{array}{rcl}\displaystyle {f(x)}&=&\displaystyle {\int f'(x)~dx}\\&&\\&=&\displaystyle {\int 3x^{2}~dx}\\&&\\&=&\displaystyle {x^{3}+D.}\\\end{array}}$
Since  $f(0)=1,$  we have
${\begin{array}{rcl}\displaystyle {1}&=&\displaystyle {f(0)}\\&&\\&=&\displaystyle {(0)^{3}+D}\\&&\\&=&\displaystyle {D.}\\\end{array}}$
Hence,
$f(x)=x^{3}+1.$

ExpandFinal Answer:
$f(x)=x^{3}+1$

Return to Sample Exam