Question

Let the functions f: R→R and g: R→R be defined as:

f(x) = `[((x + 2",", x < 0)),((x^2",", x ≥ 0))]` and

g(x) = `{{:(x^3",", x < 1),(3x - 2",", x ≥ 1):}`

Then, the number of points in R where (fog)(x) is NOT differentiable is equal to ______.

विकल्प

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Answer 1

Let the functions f: R→R and g: R→R be defined as:

f(x) = `[((x + 2",", x < 0)),((x^2",", x ≥ 0))]` and

g(x) = `{(x^3",", x < 1),(3x - 2",", x ≥ 1):}`

Then, the number of points in R where (fog)(x) is NOT differentiable is equal to 1.

Explanation:

Given that f(x) = `{(x + 2,";", x < 0),(x^2,";", x ≥ 0):}; f: R→R `

g(x) = `{(x^3,",", x < 1),(3x - 2,";", x ≥ 1):};g: R→R`

Now, f(g(x)) = `{(g(x) + 2,";", g(x) < 0),((g(x))^2,";", g(x) ≥ 0):}`

f(g(x)) = `{{:(x^3 + 2,";", x < 0),(x^6,";", 0 ≤ x ≤ 1),((3x - 2)^2,";", x ≥ 1):}`

For f(g(x)): `lim_(x→0^-)(x^3 + 2) ≠ lim_(x→0^+)(x^6)`

Hence f(g(x)) is discontinuous at x = 0.

i.e. not differentiable at x = 0.

`lim_(x→1^-)(x^6) = lim_(x →1^+)(3x - 2)^2` = 1

i.e. f(g(x)) is continuous at x = 1

So, we will check differentiability at x = 1.

R.H.D.|_x=1 = `lim_(h→0) (f(1 + h) - f(1))/h`

= `lim_(h→0) ((3(1 + h) - 2)^2 - 1)/h`

= `lim_(h→0) ((3h + 1)^2 - 1)/h`

= `lim_(h→0) (9h^2 + 6h + 1 - 1)/h`

= `lim_(h→0) (9h + 6)`

R.H.D.|_x=1 = 6

Now, L.H.D.|_x=1 = `lim_(h →0) (f(1 - h) - f(1))/(-h)`

= `lim_(h →0) ((1 - h)^6 - 1)/(-h)`

L.H.D.|_x=1 = 6

L.H.D.|_x=1 = R.H.D.|_x=1 = 6

Hence f(g(x)) is differentiable everywhere except one point, i.e., 0.

So, Number of points of non-differentiability = 1.