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# F(x) = (x^2)(e^x) on what interval is f concave down????????????????????…?

thanks =)

### 2 Answers

- 1 decade agoFavorite Answer
f'(x) = (2x)(e^x) + (x^2)(e^x) = (e^x)(2x + x^2)

f''(x) = (e^x)(2x + x^2) + (e^x)(2 + 2x) = (e^x)(2x + x^2 + 2 + 2x)

= (e^x)(x^2 + 4x + 2)

f is concave down when f'' is negative. Since e^x is always positive, then we want (x^2 + 4x + 2) < 0.

In completing the square, we get x^2 + 4x + 2 + 2 - 2 = x^2 + 4x + 4 - 2 = (x + 2)^2 - 2. So

(x + 2)^2 - 2 < 0 -> (x + 2)^2 < 2 -> |x + 2| < sqrt(2).

So f is concave down when -sqrt(2) - 2 < x < sqrt(2) - 2.

In interval notation, this is (-sqrt(2) - 2, sqrt(2) - 2).

- s kLv 71 decade ago
f'(x)=2xe^x+(x^2)e^x

f''(x)=2e^x+2xe^x+2xe^x+(x^2)e^x

0=2e^x+4xe^x+(x^2)e^x

0=(e^x)(2+4x+x^2)

0=x^2+4x+2

x=(-b+/-sqrt(b^2-4ac))/2a

=(-4+/-sqrt(16-8))/2

=(-4+/-sqrt(8))/2

=(-4+/-2sqrt(2))/2

=-2+/-sqrt(2)

=-.59,-3.41

These are the inflection points. To find where the curve is concave down, pick points around the inflection points (one less than -3.41, one between -3.41 and -.59, and one greater than -.59) and plug them into the second derivative. Whichever interval yields a negative number is concave down.

Let's use -10, -1, and 0.

f"(-10)=2e^(-10)+4(-10)e^(-10)+((-10)^2)e^(-10)

=2/(e^10)-40/(e^10)+100/(e^10)

=(102-40)/(e^10)

=82/e^10

=positive

f''(-1)=2e^(-1)+4(-1)e^(-1)+((-1)^2)e^-1

=2/e-4/e+1/e

=(3-4)/e=-1/e

=negative

f''(0)=2e^0+4(0)e^0+(0^2)e^0

=1

=positive

The curve is concave down on the interval (-3.41,-.59) (or for more accuracy (-2-sqrt(2),-2+sqrt(2))) because the second derivative is negative between the two inflection points.