Question 960391
For 
{{{f(x)=3x^2+5}}}
{{{g(x)=7x-2}}}


a. Verify:

{{{g(x + 2) <>g(x) + g(2)}}}
 
first find {{{g(x+2)}}}

{{{g(x+2)=7(x+2)-2}}}
{{{g(x+2)=7x+14-2}}}
{{{g(x+2)=7x+12}}}

and {{{g(2)}}}

{{{g(2)=7(2)-2}}}
{{{g(2)=14-2}}}
{{{g(2)=12}}}

substitute it in
{{{g(x + 2) <> g(x) + g(2)}}}
{{{7x+12 <>7x+12 + 12}}}
{{{7x+12 <>7x+24}}}



b. Find {{{(f-g)(x) }}}

since
{{{(f-g)(x) =f(x) -g(x) }}}, we have
{{{(f-g)(x) =3x^2+5 -(7x-2) }}}
{{{(f-g)(x) =3x^2+5 -7x+2}}} 
{{{(f-g)(x) =3x^2 -7x+7}}}



c. Using the resulting function in (b), show that {{{(f-g)(2) = f(2)-g(2)}}}.
(The work should be different for each side of the equation.)

{{{(f-g)(x) = f(2) – g(2)}}}
{{{3*2^2 -7*2+7= 3*2^2+5 -(7*2-2) }}}
{{{3*4 -14+7= 3*4+5 -(14-2) }}}
{{{12 -7= 12+5 -(12) }}}
{{{5= 17 -12 }}}
{{{5= 5}}}



d. Is {{{(fg)(0) =(f/g)(0)}}}? Explain.

{{{(fg)(0) =(f/g)(0)}}}
{{{(3x^2+5)(7x-2) =(3x^2+5)/(7x-2) }}}
{{{(3*0^2+5)(7*0-2) =(3*0^2+5)/(7*0-2)}}} 
{{{(5)(-2) =(5)/(-2)}}} 
{{{-10 <>-5/2 }}}.....so, the product (on the left side of equation) could not be same as the quotient (on the right side of equation)



e. Find {{{(f(x+h)-f(x))/h}}} , {{{h <> 0}}}. 

{{{(f(x+h)-f(x))/h=(3(x+h)^2+5-(3x^2+5))/h}}}


{{{(f(x+h)-f(x))/h=(3(x^2+2hx+h^2)+cross(5)-3x^2-cross(5))/h}}}


{{{(f(x+h)-f(x))/h=(cross(3x^2)+6hx+3h^2-cross(3x^2))/h}}}


{{{(f(x+h)-f(x))/h=(6hx+3h^2)/h}}}


{{{(f(x+h)-f(x))/h=3cross(h)(2x+h))/cross(h)}}}


{{{(f(x+h)-f(x))/h=3(2x+h)}}}


{{{(f(x+h)-f(x))/h=6x+3h}}}