Question

F(x)=4.5x2–3x+2

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Use the limit definition to find f ′(x).

Use the alternate limit definition, limit as x approaches a of the quantity f of x minus f of a over the quantity x minus a with a = 2, to find the value of f ′(2).

Use the shortcut rules to find f ′(x).

Find where the tangent line to f(x) is horizontal.

Write the equation of the tangent line at x = 2.

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What I've done so far:
Alternate limit definition says that f'(x) = lim x->a [f(x) - f(a)]/(x-a)
Putting f(x) in above definition
we get
lim x-> 2 [ 4.5x2-3x+2 - 4.5*22+3*2-2]/(x-2)
lim x->2 [4.5x2-3x-12]/(x-2)
lim x->2 4.5*[x+12/9]*[x-2] /(x-2) as -12/9 and 2 are roots of x2-(3/4.5)x - 12/4.5
4.5*10/3 =15

Let me know if my work is correct, see attached for easier viewing.

Answer 1

your work is confusing. please consider using latex (clicking the π looking sign at the bottom)

1. means the slope at that point
not exactly sure what the limit defintion because I learned the one you call the alternate limit definition
ah, I got it

`\lim_(h \to 0) (f(x+h)-f(x))/(h)`
so

`f'(x)= \lim_(h \to 0) (4.5(x+h)^2-3(x+h)+2-(4.5x^2-3x+2))/(h)`

`f'(x)= \lim_(h \to 0) (4.5(x^2+2xh+h^2)-3x-3h+2-4.5x^2+3x-2)/(h)`

`image`

`f'(x)= \lim_(h \to 0) (9xh+4.5h^2-3h)/(h)`

`f'(x)= \lim_(h \to 0) 9x+4.5h-3`
now simply set h=0
f'(x)=9x-3



2. nuuuh, not again

`\lim_(x \to 2) (f(x)-f(2))/(x-2)`

`\lim_(x \to 2) (4.5x^2-3x+2-(4.5(2)^2-3(2)+2))/(x-2)`

`\lim_(x \to 2) (4.5x^2-3x+2-(4.5(4)-6+2))/(x-2)`

`\lim_(x \to 2) (4.5x^2-3x+2-(18-4))/(x-2)`

`\lim_(x \to 2) (4.5x^2-3x+2-(14))/(x-2)`

`\lim_(x \to 2) (4.5x^2-3x-12))/(x-2)`
nrrrgh
oh, it divides nicely because 4.5x²-3x-12=(x-2)(4.5x+6)
4.5x+6
evaluate for x=2
4.5(2)+6
9+6
15




3.
dy/dx x^m=mx^(m-1)
dy/dx (f(x)+g(x))=dy/dx f(x)+dy/dx g(x)
dy/dx c=0 where c is a constnat
those are my shortcut rules

dy/dx 4.5x^2=4.5(2)x^(2-1)=9x^1=9x
dy/dx -3x=-3(1)x^(1-1)=-3x^0=-3(1)=-3
dy/dx 2=0
f'(x)=9x-3


4.
horizontal means slope=0
f'(x)=0
9x-3=0
9x=3
x=1/3
at x=1/3
at the point (1/3,3/2)


5.
at x=2, the slope is 15
at x=2, y=14
use point slope form
y-y1=m(x-x1)
point is (x1,y1)
slope is m
y-14=15(x-2)
most of the time it's alright to leave it in this form

Answer 2

You're on the right track, let's continue and break it down further:

The function we are considering is f(x) = 4.5x^2 - 3x + 2.

First, we need to find the derivative of this function. In calculus, the derivative measures how a function changes as its input changes. Specifically, the derivative of a function at a certain point is the slope of the line tangent to the function at that point. Mathematically, the derivative of our function, let's call it f'(x) is 9x - 3.

Now we want to find when the tangent line to the function is horizontal. The tangent line to the function is horizontal at any point where the derivative of the function is zero. Therefore, we want to solve the equation f'(x) = 0. Solving for x, we get x = 0.333.

Next, we want to find the value of the derivative at the point x = 2. This is done by simply substitifuting x = 2 into our derivative function f'(x), which gives us f'(2) = 15.

Finally, we want to find the equation of the tangent line to the function at x = 2. The slope of the tangent line is given by the derivative of the function at that point, which we already calculated as m = 15. The y-intercept of the tangent line is found by substituting x = 2 into the original function f(x), which gives us b = 14. Therefore, the equation of the tangent line is y = 15x + 14.

Comparing this with the original function f(x) = 4.5x^2 - 3x + 2, we get the equation 15x + 14 = 4.5x^2 - 3x + 2.