Math  /  Calculus

QuestionFind values of AA and BB for the function f(x)={Ax3,x<2Bx2+8,x2f(x)=\left\{\begin{array}{ll}A x^{3}, & x<2 \\ B x^{2}+8, & x \geq 2\end{array}\right. to be differentiable everywhere.

Studdy Solution
Substitute B=6B = -6 into the equation A=B3A = \frac{B}{3} to find the value of AA.
A=63=2A = \frac{-6}{3} = -2The values of AA and BB that make the function differentiable everywhere are A=2A=-2 and B=6B=-6.

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord