If f(x)=2x^2+1, what is f(x) when x =3 A)1B)7C)13D)19
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X, x+1, x+2, x+3 - four consecutive integers
The integers are 169, 170, 171 and 172.
The integers are 169, 170, 171 and 172.
Answer:
the absolute maximum value is 89.96 and
the absolute minimum value is 23.173
Step-by-step explanation:
Here we have cotangent given by the following relation;
so that the expression becomes
f(t) = 9t +9/tan(t/2)
Therefore, to look for the point of local extremum, we differentiate, the expression as follows;
f'(t) = =
Equating to 0 and solving gives
Where n is an integer hence when n₁ = 0 and n₂ = 1 we have t = π/4 and t = 3π/2 respectively
Or we have by chain rule
f'(t) = 9 -(9/2)csc²(t/2)
Equating to zero gives
9 -(9/2)csc²(t/2) = 0
csc²(t/2) = 2
csc(t/2) = ±√2
The solutions are, in quadrant 1, t/2 = π/4 such that t = π/2 or
in quadrant 2 we have t/2 = π - π/4 so that t = 3π/2
We then evaluate between the given closed interval to find the absolute maximum and absolute minimum as follows;
f(x) for x = π/4, π/2, 3π/2, 7π/2
f(π/4) = 9·π/4 +9/tan(π/8) = 28.7965
f(π/2) = 9·π/2 +9/tan(π/4) = 23.137
f(3π/2) = 9·3π/2 +9/tan(3·π/4) = 33.412
f(7π/2) = 9·7π/2 +9/tan(7π/4) = 89.96
Therefore the absolute maximum value = 89.96 and
the absolute minimum value = 23.173.