How do you determine the concavity for f(x)=x4−32x2+6?

Question

Mary Herrera · Accepted Answer

Step 1The concavity of a function is the sign of its second derivative.If, in a set, it is positive, than the concavity is up, if negative the concavity is down, if it is zero, there could be an inflection point there.so,y′=4x3−64xy″=12x2−64than12x2−64&amp;gt;0⇒x2&amp;gt;6412⇒x2&amp;gt;163⇒x&amp;lt;−43∨x&amp;gt;43, or better:x&amp;lt;−433∨x&amp;gt;433. In this set the function has concavity up, in the complementary set it has concavity dawn, in±433there are two inflection points.

Jack Maxson · Accepted Answer

Step 1
Write the polynomial as a function of x.

f (x) = x^{4} - 32 x^{2} + 6

Find the inflection points.

(- \frac{4 \sqrt{3}}{3}, - \frac{1226}{9}), (\frac{4 \sqrt{3}}{3}, - \frac{1226}{9})

The domain of the expression is all real numbers except where the expression is undefined.
In this case, there is no real number that makes the expression undefined.
Interval Notation:

(- \infty, \infty)

Set-Builder Notation:

{x ∣ x \in R}

Create intervals around the inflection points and the undefined values.

(- \infty, - \frac{4 \sqrt{3}}{3}) \cup (- \frac{4 \sqrt{3}}{3}, \frac{4 \sqrt{3}}{3}) \cup (\frac{4 \sqrt{3}}{3}, \infty)

Substitute any number from the interval

(- \infty, - \frac{4 \sqrt{3}}{3})

into the second derivative and evaluate to determine the concavity.
Concave up on

(- \infty, - \frac{4 \sqrt{3}}{3})

since

f (x)

is positive
Substitute any number from the interval

(- \frac{4 \sqrt{3}}{3}, \frac{4 \sqrt{3}}{3})

into the second derivative and evaluate to determine the concavity.
Concave down on

(- \frac{4 \sqrt{3}}{3}, \frac{4 \sqrt{3}}{3})

since

f (x)

is negative
Substitute any number from the interval

(\frac{4 \sqrt{3}}{3}, \infty)

into the second derivative and evaluate to determine the concavity
Concave up on

(\frac{4 \sqrt{3}}{3}, \infty)

since

f (x)

is positive
The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.
Concave up on

(- \infty, - \frac{4 \sqrt{3}}{3})

since

f (x)

is positive
Concave down on

(- \frac{4 \sqrt{3}}{3}, \frac{4 \sqrt{3}}{3})

since

f (x)

is negative
Concave up on

(\frac{4 \sqrt{3}}{3}, \infty)

since

f (x)

is positive

How do you determine the concavity for f(x)=x^{4}-32x^{2}+6?

Answered question

Answer & Explanation

New Questions in Calculus 1