Question

Consider the function f(x)=x2e4x. f(x) has two inflection points at x = C and x = D with C≤D where C is and D is Finally for each of the following intervals, tell whether f(x) is concave up (type in CU) or concave down (type in CD). (−[infinity],C]: [C,D]: [D,[infinity])

Answer 1

Answer:

the inflection points are

$x = -0.85, -0.14$

So,

$C = -0.85 \\D = -0.14$

It is concave down at the intervals

$(-\infty , -0.85] \cup [-0.14,\infty)$

And it would be concave up at

$(-0.85 , -0.14)$

Step-by-step explanation:

Remember that to find inflection points you need to find where

$f''(x) = 0$

Since

$f(x) = x^2 e^{4x}$

Then using the product and the chain rule you have that

$f'(x) = 2e^{4x} x(2x+1)$

And then, using again the chain rule and the product rule you have that

$f''(x) = 2e^{4x} (8x^2 +8x+1)$

Therefore you have to solve the equation

$8x^2 +8x+1 = 0$

Using the quadratic equation you get that there are two solutions, so the inflection points are

$x = -0.85, -0.14$

So,  

$C = -0.85 \\D = -0.14$

Now remember that a function is concave up if the derivative is greater than zero and concave down if the derivative is less than zero. Therefor you have to solve these inequalties

$8x^2 +8x+1 < 0 \\8x^2 +8x+1 \geq 0$

And you would get that is concave down at the intervals

$(-\infty , -0.85] \cup [-0.14,\infty)$

And it would be concave up at

$(-0.85 , -0.14)$

Answer 2

The concave upward of the function is $(-\infty, - \frac{2+\sqrt{2} }{4}) U (-\frac{2-\sqrt{2} }{4}, \infty)$ and The concave downward of the function is$\frac{-2+\sqrt{2} }{4}, \frac{2-\sqrt{2} }{4}$

Data;

• f(x) = x^2e^4x

Inflection Points

f(x) = x^2 * e^4x

solving the inflection point f''(x) = 0

$f'(x) = x^2e^4^x + e^4^x*2x = 0\\f"(x) = 4x^2*4e^4^x + 4e^4^x*2x + 2x4e^4^x + 2e^4^x = 0\\f"(x) = 4^2x^2e^4^x + 16xe^4^x + 2e^4^x = 0\\f"(x) = 2e^4^x(8x^2 + 8x + 1) = 0\\f"(x) = 8x^2 + 8x + 1 = 0$

Solving the quadatic equation above

$x = \frac{-2-\sqrt{2} }{4}, \frac{-2+\sqrt{2} }{4}$

Substituting the values into the inflection points

$(\frac{-2-\sqrt{2} }{4}, e^2^-^\sqrt{2} * \frac{3-2\sqrt{2} }{16} )$ and

$(-\frac{2+\sqrt{2} }{4}, e^2^-^\sqrt{2} * \frac{2\sqrt{3}+3 }{16} )$

The concave upward of the function is $(-\infty, - \frac{2+\sqrt{2} }{4}) U (-\frac{2-\sqrt{2} }{4}, \infty)$

The concave downward of the function is$\frac{-2+\sqrt{2} }{4}, \frac{2-\sqrt{2} }{4}$

Learn more on inflection points on a function here;

https://brainly.com/question/10352137