Find the Inflection Points 3x^4+16x^3

$3 x^{4} + 16 x^{3}$

Step 1

Write $3 x^{4} + 16 x^{3}$ as a function.

$f (x) = 3 x^{4} + 16 x^{3}$

Step 2

Find the second derivative.

Find the first derivative.

By the Sum Rule, the derivative of $3 x^{4} + 16 x^{3}$ with respect to $x$ is $\frac{d}{d x} [3 x^{4}] + \frac{d}{d x} [16 x^{3}]$ .

$f' (x) = \frac{d}{d x} (3 x^{4}) + \frac{d}{d x} (16 x^{3})$

Evaluate $\frac{d}{d x} [3 x^{4}]$ .

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{4}$ with respect to $x$ is $3 \frac{d}{d x} [x^{4}]$ .

$f' (x) = 3 \frac{d}{d x} (x^{4}) + \frac{d}{d x} (16 x^{3})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f' (x) = 3 (4 x^{3}) + \frac{d}{d x} (16 x^{3})$

Multiply $4$ by $3$ .

$f' (x) = 12 x^{3} + \frac{d}{d x} (16 x^{3})$

Evaluate $\frac{d}{d x} [16 x^{3}]$ .

Since $16$ is constant with respect to $x$ , the derivative of $16 x^{3}$ with respect to $x$ is $16 \frac{d}{d x} [x^{3}]$ .

$f' (x) = 12 x^{3} + 16 \frac{d}{d x} (x^{3})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f' (x) = 12 x^{3} + 16 (3 x^{2})$

Multiply $3$ by $16$ .

$f' (x) = 12 x^{3} + 48 x^{2}$

Find the second derivative.

By the Sum Rule, the derivative of $12 x^{3} + 48 x^{2}$ with respect to $x$ is $\frac{d}{d x} [12 x^{3}] + \frac{d}{d x} [48 x^{2}]$ .

$f'' (x) = \frac{d}{d x} (12 x^{3}) + \frac{d}{d x} (48 x^{2})$

Evaluate $\frac{d}{d x} [12 x^{3}]$ .

Since $12$ is constant with respect to $x$ , the derivative of $12 x^{3}$ with respect to $x$ is $12 \frac{d}{d x} [x^{3}]$ .

$f'' (x) = 12 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (48 x^{2})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f'' (x) = 12 (3 x^{2}) + \frac{d}{d x} (48 x^{2})$

Multiply $3$ by $12$ .

$f'' (x) = 36 x^{2} + \frac{d}{d x} (48 x^{2})$

Evaluate $\frac{d}{d x} [48 x^{2}]$ .

Since $48$ is constant with respect to $x$ , the derivative of $48 x^{2}$ with respect to $x$ is $48 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 36 x^{2} + 48 \frac{d}{d x} (x^{2})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = 36 x^{2} + 48 (2 x)$

Multiply $2$ by $48$ .

$f'' (x) = 36 x^{2} + 96 x$

The second derivative of $f (x)$ with respect to $x$ is $36 x^{2} + 96 x$ .

$36 x^{2} + 96 x$

Step 3

Set the second derivative equal to $0$ then solve the equation $36 x^{2} + 96 x = 0$ .

Set the second derivative equal to $0$ .

$36 x^{2} + 96 x = 0$

Factor $12 x$ out of $36 x^{2} + 96 x$ .

Factor $12 x$ out of $36 x^{2}$ .

$12 x (3 x) + 96 x = 0$

Factor $12 x$ out of $96 x$ .

$12 x (3 x) + 12 x (8) = 0$

Factor $12 x$ out of $12 x (3 x) + 12 x (8)$ .

$12 x (3 x + 8) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$3 x + 8 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $3 x + 8$ equal to $0$ and solve for $x$ .

Set $3 x + 8$ equal to $0$ .

$3 x + 8 = 0$

Solve $3 x + 8 = 0$ for $x$ .

Subtract $8$ from both sides of the equation.

$3 x = - 8$

Divide each term in $3 x = - 8$ by $3$ and simplify.

Divide each term in $3 x = - 8$ by $3$ .

$\frac{3 x}{3} = \frac{- 8}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 8}{3}$

Divide $x$ by $1$ .

$x = \frac{- 8}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{8}{3}$

The final solution is all the values that make $12 x (3 x + 8) = 0$ true.

$x = 0, - \frac{8}{3}$

Step 4

Find the points where the second derivative is $0$ .

Substitute $0$ in $f (x) = 3 x^{4} + 16 x^{3}$ to find the value of $y$ .

Replace the variable $x$ with $0$ in the expression.

$f (0) = 3 {(0)}^{4} + 16 {(0)}^{3}$

Simplify the result.

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$f (0) = 3 \cdot 0 + 16 {(0)}^{3}$

Multiply $3$ by $0$ .

$f (0) = 0 + 16 {(0)}^{3}$

Raising $0$ to any positive power yields $0$ .

$f (0) = 0 + 16 \cdot 0$

Multiply $16$ by $0$ .

$f (0) = 0 + 0$

Add $0$ and $0$ .

$f (0) = 0$

The final answer is $0$ .

$0$

The point found by substituting $0$ in $f (x) = 3 x^{4} + 16 x^{3}$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

Substitute $- \frac{8}{3}$ in $f (x) = 3 x^{4} + 16 x^{3}$ to find the value of $y$ .

Replace the variable $x$ with $- \frac{8}{3}$ in the expression.

$f (- \frac{8}{3}) = 3 {(- \frac{8}{3})}^{4} + 16 {(- \frac{8}{3})}^{3}$

Simplify the result.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{8}{3}$ .

$f (- \frac{8}{3}) = 3 ({(- 1)}^{4} {(\frac{8}{3})}^{4}) + 16 {(- \frac{8}{3})}^{3}$

Apply the product rule to $\frac{8}{3}$ .

$f (- \frac{8}{3}) = 3 ({(- 1)}^{4} (\frac{8^{4}}{3^{4}})) + 16 {(- \frac{8}{3})}^{3}$

Raise $- 1$ to the power of $4$ .

$f (- \frac{8}{3}) = 3 (1 (\frac{8^{4}}{3^{4}})) + 16 {(- \frac{8}{3})}^{3}$

Multiply $\frac{8^{4}}{3^{4}}$ by $1$ .

$f (- \frac{8}{3}) = 3 (\frac{8^{4}}{3^{4}}) + 16 {(- \frac{8}{3})}^{3}$

Raise $8$ to the power of $4$ .

$f (- \frac{8}{3}) = 3 (\frac{4096}{3^{4}}) + 16 {(- \frac{8}{3})}^{3}$

Raise $3$ to the power of $4$ .

$f (- \frac{8}{3}) = 3 (\frac{4096}{81}) + 16 {(- \frac{8}{3})}^{3}$

Cancel the common factor of $3$ .

Factor $3$ out of $81$ .

$f (- \frac{8}{3}) = 3 (\frac{4096}{3 (27)}) + 16 {(- \frac{8}{3})}^{3}$

Cancel the common factor.

$f (- \frac{8}{3}) = 3 (\frac{4096}{3 \cdot 27}) + 16 {(- \frac{8}{3})}^{3}$

Rewrite the expression.

$f (- \frac{8}{3}) = \frac{4096}{27} + 16 {(- \frac{8}{3})}^{3}$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{8}{3}$ .

$f (- \frac{8}{3}) = \frac{4096}{27} + 16 ({(- 1)}^{3} {(\frac{8}{3})}^{3})$

Apply the product rule to $\frac{8}{3}$ .

$f (- \frac{8}{3}) = \frac{4096}{27} + 16 ({(- 1)}^{3} (\frac{8^{3}}{3^{3}}))$

Raise $- 1$ to the power of $3$ .

$f (- \frac{8}{3}) = \frac{4096}{27} + 16 (- \frac{8^{3}}{3^{3}})$

Raise $8$ to the power of $3$ .

$f (- \frac{8}{3}) = \frac{4096}{27} + 16 (- \frac{512}{3^{3}})$

Raise $3$ to the power of $3$ .

$f (- \frac{8}{3}) = \frac{4096}{27} + 16 (- \frac{512}{27})$

Multiply $16 (- \frac{512}{27})$ .

Multiply $- 1$ by $16$ .

$f (- \frac{8}{3}) = \frac{4096}{27} - 16 (\frac{512}{27})$

Combine $- 16$ and $\frac{512}{27}$ .

$f (- \frac{8}{3}) = \frac{4096}{27} + \frac{- 16 \cdot 512}{27}$

Multiply $- 16$ by $512$ .

$f (- \frac{8}{3}) = \frac{4096}{27} + \frac{- 8192}{27}$

Move the negative in front of the fraction.

$f (- \frac{8}{3}) = \frac{4096}{27} - \frac{8192}{27}$

Combine fractions.

Combine the numerators over the common denominator.

$f (- \frac{8}{3}) = \frac{4096 - 8192}{27}$

Simplify the expression.

Subtract $8192$ from $4096$ .

$f (- \frac{8}{3}) = \frac{- 4096}{27}$

Move the negative in front of the fraction.

$f (- \frac{8}{3}) = - \frac{4096}{27}$

The final answer is $- \frac{4096}{27}$ .

$- \frac{4096}{27}$

The point found by substituting $- \frac{8}{3}$ in $f (x) = 3 x^{4} + 16 x^{3}$ is $(- \frac{8}{3}, - \frac{4096}{27})$ . This point can be an inflection point.

$(- \frac{8}{3}, - \frac{4096}{27})$

Determine the points that could be inflection points.

$(0, 0), (- \frac{8}{3}, - \frac{4096}{27})$

Step 5

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, - \frac{8}{3}) \cup (- \frac{8}{3}, 0) \cup (0, \infty)$

Step 6

Substitute a value from the interval $(- \infty, - \frac{8}{3})$ into the second derivative to determine if it is increasing or decreasing.

Replace the variable $x$ with $- 2.7\overline{6}$ in the expression.

$f'' (- 2.7\overline{6}) = 36 {(- 2.7\overline{6})}^{2} + 96 (- 2.7\overline{6})$

Simplify the result.

Simplify each term.

Raise $- 2.7\overline{6}$ to the power of $2$ .

$f'' (- 2.7\overline{6}) = 36 \cdot 7.65\overline{4} + 96 (- 2.7\overline{6})$

Multiply $36$ by $7.65\overline{4}$ .

$f'' (- 2.7\overline{6}) = 275.56 + 96 (- 2.7\overline{6})$

Multiply $96$ by $- 2.7\overline{6}$ .

$f'' (- 2.7\overline{6}) = 275.56 - 265.6$

Subtract $265.6$ from $275.56$ .

$f'' (- 2.7\overline{6}) = 9.96$

The final answer is $9.96$ .

$9.96$

At $- 2.7\overline{6}$ , the second derivative is $9.96$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, - \frac{8}{3})$ .

Increasing on $(- \infty, - \frac{8}{3})$ since $f'' (x) > 0$

Step 7

Substitute a value from the interval $(- \frac{8}{3}, 0)$ into the second derivative to determine if it is increasing or decreasing.

Replace the variable $x$ with $- 1.\overline{3}$ in the expression.

$f'' (- 1.\overline{3}) = 36 {(- 1.\overline{3})}^{2} + 96 (- 1.\overline{3})$

Simplify the result.

Simplify each term.

Raise $- 1.\overline{3}$ to the power of $2$ .

$f'' (- 1.\overline{3}) = 36 \cdot 1.\overline{7} + 96 (- 1.\overline{3})$

Multiply $36$ by $1.\overline{7}$ .

$f'' (- 1.\overline{3}) = 64 + 96 (- 1.\overline{3})$

Multiply $96$ by $- 1.\overline{3}$ .

$f'' (- 1.\overline{3}) = 64 - 128$

Subtract $128$ from $64$ .

$f'' (- 1.\overline{3}) = - 64$

The final answer is $- 64$ .

$- 64$

At $- 1.\overline{3}$ , the second derivative is $- 64$ . Since this is negative, the second derivative is decreasing on the interval $(- \frac{8}{3}, 0)$

Decreasing on $(- \frac{8}{3}, 0)$ since $f'' (x) < 0$

Step 8

Substitute a value from the interval $(0, \infty)$ into the second derivative to determine if it is increasing or decreasing.

Replace the variable $x$ with $0.1$ in the expression.

$f'' (0.1) = 36 {(0.1)}^{2} + 96 (0.1)$

Simplify the result.

Simplify each term.

Raise $0.1$ to the power of $2$ .

$f'' (0.1) = 36 \cdot 0.01 + 96 (0.1)$

Multiply $36$ by $0.01$ .

$f'' (0.1) = 0.36 + 96 (0.1)$

Multiply $96$ by $0.1$ .

$f'' (0.1) = 0.36 + 9.6$

Add $0.36$ and $9.6$ .

$f'' (0.1) = 9.96$

The final answer is $9.96$ .

$9.96$

At $0.1$ , the second derivative is $9.96$ . Since this is positive, the second derivative is increasing on the interval $(0, \infty)$ .

Increasing on $(0, \infty)$ since $f'' (x) > 0$

Step 9

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection points in this case are $(- \frac{8}{3}, - \frac{4096}{27}), (0, 0)$ .

$(- \frac{8}{3}, - \frac{4096}{27}), (0, 0)$

Step 10

Find the Inflection Points 3x^4+16x^3

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