differentiation-8

As the polynomial functions are continuous and differentiable every where f {f(x)} is continuous and differentiable in [0, 3] except perhaps at the turning points of definition, namely, x = 1, 2 denote f {f(x)} by g(x).

Now g (1+0) = 2-1=1

g (1-0) = 2+1 = 3

So g(x) is not continuous at x=1.

g (2+0) = 4-2 =2

g (2-0) = 2-2 = 0

So, g(x) is not continuous at x=2.

Hence g(x) is also not differentiable at x = 1, 2.

\The points of discontinuity and non-differentiability are x=1, 2.

Now we have to draw the graph for y = f {f(x)} where

y = 2+x in [0, 1],

y = 2-x in [1, 2]

y = 4-x in [2, 3]

The graph is discontinuous at x = 1, 2.

Fig (6)

Dumb Question:

1) In the above solution, there is a condition 0 £1+x £2 if 0 £ x £ 2 from that it is inferred that 0 £ x£ 1 How?

Ans:

0 £1+x £2 if 0 £ x £ 2

Þ -1 £ x £ 1 if 0 £ x £ 2

Now clearly both the conditions need to be satisfied simultaneously. So the intersection of these two conditions will give the desired results.

i.e. 0 £ x £ 1

2) If then find from the first principle.

Solution:

Clearly the definition of the function f(x) is where h is a very small positive quantity, remains the same, so is not a turning point of definition.

Now,

(Using definition of mod)

Now

(Using L-Hospital Rule)

3) Let a+b =1, 2a²+2b² =1 and f(x) be a continuous function such that f (2+x) +f(x) = 2 for all xÎ(0, 2) and p =then find the least positive integral values of ‘a’ for which the equation ax²²-bx+c = 0 has both roots lies between p and q;