Q1. Let f(x)=3x^3-7x^2+5x+6.The maximum value of f(x) over the interval [0,2] is
Solution
f(x)=3x^3-7x^2+5x+6.
f'(x)=9x^2-14x+5=0
=9x^2-9x-5x+5=0
=9x(x-1)-5(x-1)=0
So x=1,5/9
f"(x)=18x-14
f"(1)=18(1)-14=4>0 at point 1 function is minima
f"(5/9)=18(5/9)-14=-4<0
so point 5/9 function is maxima
Value at point 5/9
f(5/9)=3(5/9)^3-7(5/9)^2+5(5/9)+6
=7.1
Value at point 0 f(0)=3(0)^3-7(0)^2+5(0)+6
=0
Value at point 2
f(2)=3(2)^3-7(2)^2+5(2)+6
=12
So maximum value is 12 at point 2.
Q2. At the point x=0 the function f(x)=x^3 has what value local maxima or minima
Solution
f(x)=x^3
f'(x)= 3x^2
At value x =0
f'(x)=0
f"(x)=6x
At point x=0
f"(x)=0
f"'(x)=6
6 not equal to 0
So saddle point exit neither maxima neither minima
Q3.The minimum value of the function f(x)=(x^3/3)-x at point point
Solution
f(x)=(x^3/3)-x
f'(x)=3x^2/3-1
f'(x)=x^2-1
So x^2=1
So x=-1,+1
f"(x)=2x
At point 1
=2>0
At point -1
=-2<0
Because 2 is greater than 0 so minimum value of function at point 1.
Q4. The function f(x)=2x^3-3x^2 in the domain [-1,2].The global minimum of f(x) is
Solution
f(x)=2x^3-3x^2
f'(x)=6x^2-6x=0
=6x(x-1)=0
=x=0,1
f"(x)=12x-6
f"(0)=-6 <0
So function is maxima at point 0
f"(1)=12-6=6>0
f(1)=2(1)^3-3(1)^2=-1
f(-1)=-2-3=-5
f(2)=16-12=4
So global minimum value is -5 at point 1 .
Q5.Let f[-1,1]->R where f(x)=2x^3-x^4-10 .The minimum value of f(x) is
Solution
f(x)=2x^3-x^4-10
f'(x)=6x^2-4x^3=0
x^2(6-4x)=0
x=0,3/2
f"(x)=12x-12x^2
f"(0)=12(0)-12(0)^2
f"(0)=0
f"'(0)=12-24x
f"'(0) value is not equal to zero so saddle point at point 0.
Point 3/2 is not in function range f[-1,1].
f(-1)=2(-1)^3-(-1)^4-10
=-13
f(1)=2(1)^3-(1)^4-10
=2-1-10
=-9
The minimum value of f(x) is -13
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