Question

I want to make sure i got this right. PLS HELP ASAP I NEED IT CALC 1 one question

Answer 1

$\bf f(x)= \begin{cases} x^3&x\le 1\\ mx+b&x\ \textgreater \ 1 \end{cases}$

now, a function may well be continuous, but it may not be differentiable.

graph wise, differentiability means, a smooth curve, with no abrupt spikes or drops or cuts.

so for this piece-wise to be differentiable, the subfunctions must meet at junctions smoothly, with no abrupt changes, so mx + b, must blend in with x³ smoothly.

now, that can only occur if mx + b has the same slope as x³ at that point where they meet.  The point where they meet is at x = 1, well, let's check what is the slope of x³ at x = 1.

$\bf \cfrac{d}{dx}[x^3]\implies \left. 3x^2\cfrac{}{}\right|_{x=1}\implies 3(1)^2\implies 3$

therefore "m" for mx+b, must be 3 then.

now, what's the value of x³ at x = 1?  well is just (1)³, which is 1.

so for 3x + b, when x = 1, it must also equals 1 as well, that way it meets x³, thus

3x + b

3(1) + b = 1

b = -2

check the graph below.