- #1
Caeder
- 13
- 0
I have to prove [tex]x^3[/tex] is differentiable at [tex]x=4[/tex] using the definition of what it means for something to be differentiable.
So I was wondering if I just have to show that [tex]f'(4) = \lim_{x \to 4} \frac {f(x) - f(4)}{x - 4}[/tex] exists, where [tex]f(x) = x^3[/tex].
So...
[tex]f'(4) = \lim_{x \to 4} \frac {x^3 - 64}{x - 4}[/tex]
[tex]= \lim_{x \to 4} \frac{(x-4)(x^2 + 4x + 16)}{x-4}[/tex]
[tex] = \lim_{x \to 4} x^2 + 4x + 16[/tex].
So the limit (derivative) is 48...does this prove it's dif'able at [tex]x=4[/tex]?
So I was wondering if I just have to show that [tex]f'(4) = \lim_{x \to 4} \frac {f(x) - f(4)}{x - 4}[/tex] exists, where [tex]f(x) = x^3[/tex].
So...
[tex]f'(4) = \lim_{x \to 4} \frac {x^3 - 64}{x - 4}[/tex]
[tex]= \lim_{x \to 4} \frac{(x-4)(x^2 + 4x + 16)}{x-4}[/tex]
[tex] = \lim_{x \to 4} x^2 + 4x + 16[/tex].
So the limit (derivative) is 48...does this prove it's dif'able at [tex]x=4[/tex]?