009B Sample Midterm 3, Problem 3 Detailed Solution

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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.

Background 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:

Step 1:
Since  $f(x)$ passes through the point  $(0,1),$ $f(0)=1.$ Since  $f(x)$ has a horizontal tangent at  $(0,1),$ $f'(0)=0.$ Step 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,
$f'(x)=3x^{2}.$ Step 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,
$f(x)=x^{3}+1.$ $f(x)=x^{3}+1$ 