1. The limit definition for solving a derivative is as follows:
#lim_(h->0)(f(x+h)-f(x))/h#
So, let's start by plugging the equation in:
#lim_(h->0)([4.5(x+h)^2-3(x+h)+2]-[4.5x^2-3x+2])/h#
Simplify:
#lim_(h->0)[color(red)(4.5x^2)+9hx+4.5h^2color(red)(-3x)-3h color(red)(+2]color(red)(-)[color(red)(4.5x^2-3x+2)])/h#
#lim_(h->0)(cancel(color(red)(4.5x^2))+9hx+4.5h^2cancel(color(red)(-3x))-3hcolor(red)(cancel(+2)cancel(-4.5x^2)cancel(+3x)cancel(-2)))/h#
#lim_(h->0)(4.5h^2+9hx-3h)/h#
Next, factor out #h# from the numerator:
#lim_(h->0)(color(red)(h)(4.5h+9x-3))/color(red)(h)#
The #h#'s now cancel:
#lim_(h->0)(cancel(color(red)(h))(4.5h+9x-3))/cancel(color(red)(h))#
Finally, apply the limit using substitution:
#lim_(h->0)4.5h+9x-3=4.5(0)+9x-3=9x-3#
#f'(x)=9x-3#
2. Plug the equation into the alternate limit definition:
#lim_(x->a)(f(x)-f(a))/(x-a)#
#lim_(x->2)([4.5x^2-3x+2]-[4.5(2)^2-3(2)+2])/(x-2)#
Simplify:
#lim_(x->2)((4.5x^2-3x+2)-(18-6+2))/(x-2)#
#lim_(x->2)(4.5x^2-3x-12)/(x-2)#
The numerator factors:
#lim_(x->2)(1.5(3x+4)color(red)((x-2)))/color(red)((x-2))#
Simplify:
#lim_(x->2)(1.5(3x+4)cancel(color(red)((x-2))))/cancel(color(red)((x-2)))#
#lim_(x->2)4.5x+6#
Finally, apply the limit using substitution:
#lim_(x->2)4.5x+6=4.5(2)+6=15#
3. The derivative rules we will use here will be the power rule, the sum rule, and the constant rule:
This is the power rule:
#d/dx(x^n)=n*x^(n-1)#
This is the sum rule:
#d/dx(u+v)=(du)/dx+(dv)/dx#
This is the constant rule, where #k# is the constant:
#d/dx(k)=0#
Let's start with the sum rule:
First, we will equate #f(x)# and #y# to make the notation simpler.
#f(x)=y#
Using the sum rule, we will differentiate each term individually:
#(dy)/dx=d/dx(4.5x^2)-d/dx(3x)+d/dx(2)#
Next, use the power rule to take the derivative of the first two terms:
#(dy)/dx=(2*4.5x^(2-1))-(1*3x^(1-1))+d/dx(2)#
By the constant rule, the derivative of any constant is 0:
#(dy)/dx=(2*4.5x^(2-1))-(1*3x^(1-1))+0#
Finally, simplify:
#(dy)/dx=9x-3#
4. The derivative of a function at a certain point gives the slope of the line tangent to that point.
Horizontal lines have a slope of zero, so we can set our derivative equal to zero, to get the x value for that point:
#9x-3=0#
#x=1/3#
Now, plug #1/3# into the original function to get its corresponding y-value.
#f(1/3)=4.5(1/3)^2-3(1/3)+2#
Simplify:
#f(1/3)=3/2#
So, we know that the function has a tangent line with a slope of zero (horizontal) at #(1/3,3/2)#.
5. We know from Part 2 that the slope of the tangent line at #x=2# is #15#.
Now, plug #2# into the original function to get its corresponding y-value.
#f(2)=4.5(2)^2-3(2)+2#
Simplify:
#f(2)=14#
Finally, plug this information into the point-slope formula:
#y_2-y_1=m(x_2-x_1)#
#y-14=15(x-2)#
This can also be written in slope intercept form if #y# is isolated:
#y-14=15x-30#
#y=15x-16#
