159k views
4 votes
Find the values of m and b that make the following function differentiable.

f(x) = {x^2, x less than or equal to 2
mx+b, x>2}
thank you so much!!

User Tom Johns
by
8.3k points

1 Answer

5 votes

Both x² and mx + b are differentiable functions of x (they are both polynomials), so if f(x) is also differentiable, we need to pay special attention at x = 2 where the two pieces of f meet.

Continuity means that the limit


\displaystyle \lim_(x\to2) f(x)

must exist.

From the left side, we have x < 2 and f(x) = x², so


\displaystyle \lim_(x\to2^-) f(x) = \lim_(x\to2) x^2 = 4

From the right, we have x > 2 and f(x) = mx + b, so


\displaystyle \lim_(x\to2^+) f(x) = \lim_(x\to2) (mx+b) = 4m+b

It follows that 4m + b = 4.

Differentiability means that the limit


\displaystyle \lim_(x\to2) (f(x) - f(2))/(x - 2)

must exist.

From the left side, we again have x < 2 and f(x) = x². Then


\displaystyle \lim_(x\to2^-)(f(x)-f(2))/(x-2) = \lim_(x\to2) (x^2-4)/(x-2) = \lim_(x\to2) (x+2) = 4

From the right side side, we have x > 2 so f(x) = mx + b. Then


\displaystyle \lim_(x\to2^+)(f(x)-f(2))/(x-2) = \lim_(x\to2) ((mx+b)-(2m+b))/(x-2) = \lim_(x\to2) (mx-2m)/(x-2) = \lim_(x\to2)m = m

The one-sided limits must be equal, so m = 4, and from the other constraint it follows that 16 + b = 4, or b = -12.

User ManishSingh
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories