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
$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
so
$f'(x)= \lim_{h \to 0} \frac{4.5(x+h)^2-3(x+h)+2-(4.5x^2-3x+2)}{h}$
$f'(x)= \lim_{h \to 0} \frac{4.5(x^2+2xh+h^2)-3x-3h+2-4.5x^2+3x-2}{h}$
$f'(x)= \lim_{h \to 0} \frac{4.5x^2+9xh+4.5h^2-3x-3h+2-4.5x^2+3x- 2}{h}$
$f'(x)= \lim_{h \to 0} \frac{9xh+4.5h^2-3h}{h}$
$f'(x)= \lim_{h \to 0} 9x+4.5h-3$
now simply set h=0
f'(x)=9x-3



2. nuuuh, not again
$\lim_{x \to 2} \frac{f(x)-f(2)}{x-2}$
$\lim_{x \to 2} \frac{4.5x^2-3x+2-(4.5(2)^2-3(2)+2)}{x-2}$
$\lim_{x \to 2} \frac{4.5x^2-3x+2-(4.5(4)-6+2)}{x-2}$
$\lim_{x \to 2} \frac{4.5x^2-3x+2-(18-4)}{x-2}$
$\lim_{x \to 2} \frac{4.5x^2-3x+2-(14)}{x-2}$
$\lim_{x \to 2} \frac{4.5x^2-3x-12)}{x-2}$
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