stuck on this one problem

**soldatik21** · March 2nd 2009, 08:47 PM

Can anyone help me out with this one problem please, didn't really learn this type of problem yet.

find a and b such that f is differentiable everywhere

f(x)= ax^3, x=less than or equal to 2
x^2+b, x=greater than 2

thanks

**Abu-Khalil** · March 2nd 2009, 09:15 PM

You only have problems to differentiate at $x=2$ so you must find $a,b$ such as $f$ is continuous and differentiable there, i.e. to find values such as $\lim_{x\to 2^{-}}f(x)=\lim_{x\to 2^{+}}f(x)=f(2)$ and $\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}$ exists.

$\lim_{x\to 2^{+}}f(x)=\lim_{x\to 2^{+}}x^2+b=4+b$ and $f(2)=8a$ so our first condition is $4+b=8a$ .

On the other hand, $\lim_{h\to 0^{+}}\frac{f(2+h)-f(2)}{h}=\lim_{h\to 0^{+}}\frac{(2+h)^2+b-4-b}{h}=\lim_{h\to 0^{+}}\frac{(2+h)^2-4}{h}=4.$ and $\lim_{h\to 0^{-}}\frac{f(2+h)-f(2)}{h}=\lim_{h\to 0^{-}}\frac{a(2+h)^3-8a}{h}=4a$ so our second condition is $4a=4$ .

Finally $a=1$ and $b=4$ .

**soldatik21** · March 2nd 2009, 09:31 PM

thanks for showing me how to do it. just one question, could there be another answer because in my text book it is different than what you gave me.

**mr fantastic** · March 3rd 2009, 02:30 AM

Originally Posted by soldatik21

Can anyone help me out with this one problem please, didn't really learn this type of problem yet.

find a and b such that f is differentiable everywhere

f(x)= ax^3, x=less than or equal to 2
x^2+b, x=greater than 2

thanks

Continuity at x = 2: a(2)^3 = (2)^2 + b => 8a = 4 + b .... (A)

Differentiability at x =2: 3a(2)^2 = 2(2) => 12a = 4 => a = 1/3.

Substitute this value of a into equation (A) to get b.

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