Please show how you did each?

enter image source here

1 Answer
Jan 19, 2018

See below. Each part is bolded.

Explanation:

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