Answer:
B
A
Step-by-step explanation:
[tex]f(x) = \left \{ \left \begin{matrix}5x-6,\ x<4\\x^{2}-2,\ 4 \leq x \leq 6\\4x+10,\ x>6 \end{matrix} \right } \right[/tex]
To find f'(x), take the derivative within each interval. Clearly, it's 5 in the first one, 2x in the second one, and 4 in the third one. But we need to determine if the derivative exists at the ends of each interval.
For a derivative to exist at a point x=a, the function must be continuous (f(a⁻) = f(a⁺)), and smooth (f'(a⁻) = f'(a⁺)).
Let's look at x=4. On the left side:
f(4⁻) = 14, f'(4⁻) = 5
On the right side:
f(4⁺) = 14, f'(4⁺) = 8
So the function is continuous, but not smooth. Therefore, the derivative doesn't exist there.
Let's try again with x=6. On the left side:
f(6) = 34, f'(6) = 12
On the right side:
f(6⁺) = 34, f'(6⁺) = 4
Again, the function is continuous, but not smooth, so the derivative doesn't exist there either.
Therefore, f'(x) is:
[tex]f'(x) = \left \{ \left \begin{matrix}5,\ x<4\\2x,\ 4<x<6\\4,\ x>6 \end{matrix} \right } \right\ Domain:\ All\ real\ numbers,\ x\neq 4,6[/tex]
When we graph f(x) = (4x² − 1) / (x² − 9), we see f(x) has a horizontal tangent line at x=0.
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