1)state te domain
2)determine the intercepts, if any
3)discuss symmetry of graph
4)find any asympotes
5)determine intervals of increase and decrease
6)what is the maxima and/or minima
7) where is the curve upward and downward
8)locate the points of inflection
I think 1. is all real numbers, 2. Is (0,2), 3. Is the graph is symmetric with respect to the yaxis, 6. Is maxima (0,2), but I'm not sure plus I'm clueless for the rest, any help out there?
2)determine the intercepts, if any
3)discuss symmetry of graph
4)find any asympotes
5)determine intervals of increase and decrease
6)what is the maxima and/or minima
7) where is the curve upward and downward
8)locate the points of inflection
I think 1. is all real numbers, 2. Is (0,2), 3. Is the graph is symmetric with respect to the yaxis, 6. Is maxima (0,2), but I'm not sure plus I'm clueless for the rest, any help out there?
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Hello,
f(x) = 2.e^(-5x²)
1.= = = = = = = = = =
Exponential function is defined on IR. Thus f is defined on all real numbers. You are correct.
2.= = = = = = = = = =
Exponential function is always positive, so f(x)=0 is impossible. So no x-intercept.
When x=0, then f(0)=2. Thus y-intercept is (0; 2). You are correct.
3.= = = = = = = = = =
Since x²=(-x)² for any real value x, f(x)=f(-x). Thus is the function even. Thus graph is symmetric with respect to y-axis. You are correct.
4.= = = = = = = = = =
Since Lim(x→±∞) f(x)=0, the line of equation y=0 (x-axis) is an horizontal asymptote of the function.
5.= = = = = = = = = =
f'(x) = -20x.e^(-5x²)
Since e^(-5x²) is always positive, the sign of f' is the opposite sign of x.
Thus f is increasing on IR⁻ and decreasing on IR⁺.
6.= = = = = = = = = =
Because of the above answer, we can conclude that the only extremum is located at x=0 and that it is a maximum.
So the maximal value of f is located at x=0 and f(0)=2. You are correct.
7.= = = = = = = = = =
f'(x) = -20x.e^(-5x²)
There is no real value such that the curve has an infinite slope.
8.= = = = = = = = = =
f"(x) = -20.e^(-5x²) + 200x².e^(-5x²)
Inflection points are obtained when f"(x) = 0
-20.e^(-5x²) + 200x².e^(-5x²) = 0
-1 + 10x² = 0
10x² = 1
x² = 1/10
x = ±(√10) / 10
f[±(√10) / 10] = 2.e^(-5/10) = 2.e^(-½)
Thus the inflexions points are:
( -(√10)/10; .e^(-½) ) and ( +(√10)/10; .e^(-½) )
Regards,
Dragon.Jade :-)
f(x) = 2.e^(-5x²)
1.= = = = = = = = = =
Exponential function is defined on IR. Thus f is defined on all real numbers. You are correct.
2.= = = = = = = = = =
Exponential function is always positive, so f(x)=0 is impossible. So no x-intercept.
When x=0, then f(0)=2. Thus y-intercept is (0; 2). You are correct.
3.= = = = = = = = = =
Since x²=(-x)² for any real value x, f(x)=f(-x). Thus is the function even. Thus graph is symmetric with respect to y-axis. You are correct.
4.= = = = = = = = = =
Since Lim(x→±∞) f(x)=0, the line of equation y=0 (x-axis) is an horizontal asymptote of the function.
5.= = = = = = = = = =
f'(x) = -20x.e^(-5x²)
Since e^(-5x²) is always positive, the sign of f' is the opposite sign of x.
Thus f is increasing on IR⁻ and decreasing on IR⁺.
6.= = = = = = = = = =
Because of the above answer, we can conclude that the only extremum is located at x=0 and that it is a maximum.
So the maximal value of f is located at x=0 and f(0)=2. You are correct.
7.= = = = = = = = = =
f'(x) = -20x.e^(-5x²)
There is no real value such that the curve has an infinite slope.
8.= = = = = = = = = =
f"(x) = -20.e^(-5x²) + 200x².e^(-5x²)
Inflection points are obtained when f"(x) = 0
-20.e^(-5x²) + 200x².e^(-5x²) = 0
-1 + 10x² = 0
10x² = 1
x² = 1/10
x = ±(√10) / 10
f[±(√10) / 10] = 2.e^(-5/10) = 2.e^(-½)
Thus the inflexions points are:
( -(√10)/10; .e^(-½) ) and ( +(√10)/10; .e^(-½) )
Regards,
Dragon.Jade :-)