Answer to Question #114620 in Calculus for ANJU JAYACHANDRAN

Question

Let f(x)={3x^2, x lessthan or equal to1;ax+b, x graterthan 1
Find the value of a and b so that f is differentiable at x=1.

Answer 1

$f(x) = 3 x^2$ when $x \leq 1.$ $\implies f(1)=3$

$f(x)=ax+b$ when $x>1 \implies \lim_{x \to 1+}f(x)=a+b$

A differentiable function is also continuous. The function $f(x)$ is continuous at $x=1$ iff $\lim_{x \to 1-}f(x)=\lim_{x \to 1+}f(x)=f(1),$ hence $a+b-3=0.$

Now,

$f'(1^-)= \lim_{h \to 0} \frac{f(1-h)-f(1)}{-h} = \lim_{h \to 0} \frac{3(1-h)^2-3}{-h} =\\= \lim_{h \to 0} \frac{3h^2-6h}{-h} = \lim_{h \to 0} 6-3h = 6$

$f'(1^+) = \lim_{h \to 0} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0} \frac{a(1+h)+b-3}{h} =\\= \lim_{h \to 0} \frac{a+b-3+ah}{h} = a$

The functuon is differentiable iff $f'(1^-)=f'(1^+) \implies a=6$.

From $a=6$ and $a+b-3=0$ it follows that $a=6, b=-3$ .