Question

Help Please!
Find the values of m and b that make the following function differentiable.
(This is a piecewise function.)
f(x)= {x^3, x≤1
{mx+b, x>1

Answer 1

what is the value of x^3 when x = 1?

Answer 2

3

Answer 3

x^3 doesn't mean x times 3

Answer 4

1 lol

Answer 5

srry I didn't read that right

Answer 6

yep x^3 is equal to 1 when x = 1

Answer 7

what is mx+b equal to when x = 1 ?

Answer 8

1=m(1)+b

Answer 9

so you should find that 1 = m+b

solve for m or b. I'm going to solve for b to get

b = 1-m

Answer 10

now let's derive each piece

x^3 derives to 3x^2
mx+b derives to just m

agreed? or no?

Answer 11

Yes agreed

Answer 12

ok for the function f(x) to be differentiable at x = 1, f ' (x) needs to be continuous at x = 1

so we need to plug x = 1 into each piece (3x^2 and m) and set those outputs equal to each other

first piece: 3x^2 ----> 3(1)^2 = 3
second piece: m ----> m ... there are no x values to plug in x = 1

so we know that m = 3

Answer 13

so it's 3x^2= 3(1)+b so far

Answer 14

if m = 3 and we know b = 1-m, then what is b?

Answer 15

-2=b

Answer 16

good

Answer 17

Thank you :)

Answer 18

no problem