Application of DerivativeMedium
Question
If f(x) 
, then
, thenOptions
A.f(x) is increasing on [-1, 2]
B.f(x) is continuous on [-1,3]
C.f′(2) does not exist
D.f(x) has the maximum value at x = 2
Solution
For - 1 ≤ x ≤ 2 we have
f(x) = 3x2 + 12x - 1
⇒ f′(x) = 6x + 12 > 0, ∀ - 1 ≤ x ≤ 2
Hence, f(x) is increasing in [-1, 2]
A gain function is a algebraic polynomial, therefore, it is continuous at ∈ (-1, 2) and 2, 3) and continuity at x = 2
f(x) (3x2 + 12x - 1)
[3(2 - h)2 + 12(2 - h) - 1]
[3 (4 + h2 - 4h) + 24 - 12h - 1]
(12 + 3h2 - 12h + 24 - 12h - 1)
(3h2 - 24h + 35) = 35
and
f(x) (37 - x)
[37 - (2 + h)] = 35
and f(2) = 3.22 +12.2 -1-12 + 24 -1 = 35
Therefoer, LHL = RHL = f(2) ⇒ function is continuous at x = 2 ⇒ function is continuous in - 1 ≤ x ≤3,
Now, Rf′(2)
and Lf′(2)
Since, Rf′(2) ≠ Lf′(2), fR
f(x) = 3x2 + 12x - 1
⇒ f′(x) = 6x + 12 > 0, ∀ - 1 ≤ x ≤ 2
Hence, f(x) is increasing in [-1, 2]
A gain function is a algebraic polynomial, therefore, it is continuous at ∈ (-1, 2) and 2, 3) and continuity at x = 2
f(x) (3x2 + 12x - 1) [3(2 - h)2 + 12(2 - h) - 1] [3 (4 + h2 - 4h) + 24 - 12h - 1] (12 + 3h2 - 12h + 24 - 12h - 1) (3h2 - 24h + 35) = 35and
f(x) (37 - x) [37 - (2 + h)] = 35 and f(2) = 3.22 +12.2 -1-12 + 24 -1 = 35
Therefoer, LHL = RHL = f(2) ⇒ function is continuous at x = 2 ⇒ function is continuous in - 1 ≤ x ≤3,
Now, Rf′(2)
and Lf′(2)
Since, Rf′(2) ≠ Lf′(2), fR
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