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**HallsofIvy** Well, it helps to know what "continuous" means! A function, f(x), is said to be "continuous at x= a" if and only if $\lim_{x\to a} f(x)= f(a)$. Taking the limit of f(x) as x approaches 0 from below, $lim_{x\to 0} 4x= 0$. Taking the limit of f(x) as x approaches 0 from above, $\lim_{x\to 0} kx+ 2kx^2= 0$. So the limits is 0, for any k. Further, f(0)= 0. This function is continuous at 0 for **all** k. And, each part is a polynomial so it is necessarily continuous for x positive or negative.

This function is continuous for **all** values of k.

For the second part, you need to know what "differentiable" means! The "derivative of f at x= a" is $\lim_{h\to 0}\frac{f(a+ h)- f(a)}{h}$. A function **is** "differentiable" if and only if that derivative exists. Again, every polynomial is differentiable for all x so f is clearly differentiable for all x other than 0. To find the derivative at x= 0, take the "limit from below" $\lim_{h\to 0}\frac{4(x+ h)- 4x}{h}= \lim_{h\to 0} \frac{4x+ 4h- 4x}{h}= \lim_{h\to 0} \frac{4h}{h}= 4$ and then take the "limit from above" $\lim_{h\to 0} \frac{k(x+h)+ 2k(x+ h)^2- kx- 2kx^2}{h}= \lim_{h\to 0}\frac{kx+ kh+ 2kx^2+ 4khx+ kh^2- kx- 2x^2}{h}= \lim_{h\to 0}\frac{kh+ 4khx+ kh^2}{h}= \lim_{h\to 0} k+ 4kx+ kh= k$.

In order that f be differentiable at x= 0, those two limits must be the same: k= 4.