Answer to Question #131750 in Calculus for Navya Sharma

Question

let f(x)={3x^2,x<=1}{ax+b,x>1} Find the value of a and b so that f is differentiable at x =1

Answer 1

$f(x) = \begin{cases} 3x^{2} &\text{if } x<=1 \\ ax+b &\text{if } x>1 \end{cases} \\\lim\limits_{x\mapsto1}ax+b=3\implies a+b=3 \\ \tfrac{d}{dx}(3x^{2})\mid_{x=1}=\tfrac{d}{dx}(ax+b)\mid_{x=1}\implies a=6\implies b=-3 \\Answer:a=6, b=-3$