Question

Let f be the function defined below where c and d are constant. If f is differentiable at x=-2, what is the value of c+d?

Answer 1

lol

Answer 2

First, the definition of \(f(x)\) looks incorrect, since you appear to define \(f(-2)\) twice: The upper definition you gave is for \(x \le -2\) and the lower definition is for \(x \ge -2\). Either the upper definition should be for \(x < -2\) or the lower definition should be for \(x > -2\).

Now, having said that, the key to solving the problem is the statement that \(f(x)\) is differentiable at \(x=-2\). Clearly \(f(x)\) is differentiable in the interval below -2 (because it's defined as a simple linear function of \(x\) on that interval), and it's also differentiable on the interval above -2 (because it's defined as a quadratic equation in \(x\) on that other interval). So you have two derivatives: \(f'(x)\) for \(x < -2\) and \(f'(x)\) for \(x > -2\). (I'll leave it to you to compute what the derivatives are.)

If \(f(x)\) is differentiable at \(x=-2\) also then \(f'(2)\) exists and it's equal to the limit of \(f'(x)\) as \(x\) approaches -2 from below and above: \[\lim_{x\rightarrow-2^-} f'(x) = \lim_{x\rightarrow-2^+} f'(x) = f'(2)\]So basically you take the two derivatives, see what their values would be at \(x=-2\), and equate them. That gives you an equation involving \(c\) and \(d\).

You can get a second equation involving \(c\) and \(d\) by noting that if \(f(x)\) is differentiable at \(x=-2\) then it's also continuous at \(x=-2\). In other words,\[\lim_{x\rightarrow-2} f(x) = f(-2)\]So again you can take the limits from above and below, using the definition of \(f(x)\), and equate the limits. This will give you another equation involving \(c\) and \(d\).

At that point you'll have two equations in two unknowns, and you can solve for \(c\) and \(d\). I hope this gives you enough hints to solve the problem.