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Find values of x where the tangent line is horizontal for the graph of fx equals 4x2 divided by x plus 2?
Find values of x that make the tangent line to
f(x)=4x2/(x+2)
horizontal.
For a tangent line to be horizontal, its slope must be zero. The first derivative of a function will give the slope of the tangent line at any point on that function. So, by setting the first derivative of the function equal to zero and solving for x, we can find the values of x that will achieve the desired outcome.
First, to derive f(x):
Traditionally, you would use the quotient rule to derive this. If you know how to do so and prefer using the quotient rule, go ahead and do so, ignoring the next few steps. I personally don't remember the quotient rule because I always use the product rule. Essentially, I see f(x) as:
f(x)=(4x2)*(1/(x+2))
and derive it using the product rule. Doing this, I get the rough, unsimplified first derivative:
f'(x)=(4x2)*-(x+2)-2*(1)+(8x)*(x+2)-1
f'(x)=-4x2/(x+2)2+8x/(x+2)
I can add these two fractions by using a common denominator of (x+2)2:
f'(x)=-4x2/(x+2)2+(8x*(x+2))/(x+2)2
f'(x)=-4x2/(x+2)2+(8x2+16x)/(x+2)2
f'(x)=(8x2-4x2+16x)/(x+2)2
f'(x)=(4x2+16x)/(x+2)2
This final simplification is a good simplified first derivative of f(x). We must now find what x-values will cause this first derivative to equal zero. Since it is a fraction with a numerator of 4x2+16x and a denominator of (x+2)2, it will equal zero whenever the numerator is equal to zero. So, by setting just the numerator equal to zero, we get:
4x2+16x=0
Factor out as much as you can...
4x(x+4)=0
Since there are two components that are multiplied together to equal zero, setting either to zero will yield a valid solution to the above equation. So, set each equal to zero individually to get all valid values of x for your problem:
4x=0
x=0/4
x=0
and
x+4=0
x=0-4
x=-4
So, in conclusion, the two values of x that will yield horizontal tangent lines for f(x) are x=0 and x=-4 because these values will make the first derivative of f(x) equal zero.
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