For what values of x is #f(x)=(2x^2−4x)e^x# concave or convex?

Answer 1

The function is convex for #x in (-oo,-1-sqrt3) uu(-1+sqrt3, +oo)# and concave for #x in (-1-sqrt3, -1+sqrt3)#

#"Reminder"#
#(uv)'=u'v+uv'#

Calculate the first and second derivatives

#f(x)=(2x^2-4x)e^x=2(x^2-2x)e^x#
#f'(x)=2*(2x-2)e^x+2(x^2-2x)e^x=2(x^2-2)e^x#
#f''(x)=2*(2x)e^x+2(x^2-2)e^x=2(x^2+2x-2)e^x#
The inflection points are when #f''(x)=0#
#x^2+2x-2=0#
Solving this quadratic equation for #x#
#x=(-2+-sqrt(2^2-4*1*(-2)))/(2)=(-2+-sqrt(12))/(2)#
#=(-2+-2sqrt(3))/2#
#=-1+-sqrt3#

The roots are

#x_1=-1-sqrt3#
#x_2=-1+sqrt3#

We can build the variation chart

#color(white)(aaa)##color(white)(aaa)##"Interval"##color(white)(aaa)##(-oo, x_1)##color(white)(aaa)##(x_1, x_2)##color(white)(aaa)##(x_2, +oo)#
#color(white)(aaa)##color(white)(aaa)##"Sign f''(x)"##color(white)(aaaaa)##+##color(white)(aaaaaaaa)##-##color(white)(aaaaaaa)##+#
#color(white)(aaa)##color(white)(aaa)##" f(x)"##color(white)(aaaaaaaaa)##uu##color(white)(aaaaaaaa)##nn##color(white)(aaaaaaa)##uu#

graph{(2x^2-4x)e^x [-17.35, 14.68, -7.95, 8.07]}

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Answer 2
To determine the concavity or convexity of the function \( f(x) = (2x^2 - 4x)e^x \), we need to find the second derivative and examine its sign. First derivative of \( f(x) \): \[ f'(x) = (4x - 4)e^x + (2x^2 - 4x)e^x = (2x^2 + 4x - 4)e^x \] Second derivative of \( f(x) \): \[ f''(x) = (4x + 4)e^x + (2x^2 + 4x - 4)e^x = (2x^2 + 8x)e^x \] To determine concavity or convexity, we look at the sign of the second derivative. - If \( f''(x) > 0 \), the function is convex. - If \( f''(x) < 0 \), the function is concave. Therefore, \( f(x) \) is: - Convex when \( (2x^2 + 8x)e^x > 0 \). - Concave when \( (2x^2 + 8x)e^x < 0 \). Since \( e^x \) is always positive, we only need to consider the sign of \( (2x^2 + 8x) \). \( (2x^2 + 8x) \) can be factored as \( 2x(x + 4) \). The sign of \( (2x^2 + 8x) \) depends on the sign of \( x \) and \( (x + 4) \): - If \( x < -4 \), both \( x \) and \( (x + 4) \) are negative, so \( (2x^2 + 8x) > 0 \), making the function convex. - If \( -4 < x < 0 \), \( x \) is negative but \( (x + 4) \) is positive, so \( (2x^2 + 8x) < 0 \), making the function concave. - If \( x > 0 \), both \( x \) and \( (x + 4) \) are positive, so \( (2x^2 + 8x) > 0 \), making the function convex. Therefore, \( f(x) = (2x^2 - 4x)e^x \) is convex for \( x < -4 \) and \( x > 0 \), and concave for \( -4 < x < 0 \).
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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