Calculus Examples

f (x) = 0.0000329 t^{3} - 0.0061 t^{2} + 0.0514 t + 417

$f (x) = 0.0000329 t^{3} - 0.0061 t^{2} + 0.0514 t + 417$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

By the Sum Rule, the derivative of

0.0000329 t^{3} - 0.0061 t^{2} + 0.0514 t + 417

$0.0000329 t^{3} - 0.0061 t^{2} + 0.0514 t + 417$ with respect to

x

$x$ is

\frac{d}{d x} [0.0000329 t^{3}] + \frac{d}{d x} [- 0.0061 t^{2}] + \frac{d}{d x} [0.0514 t] + \frac{d}{d x} [417]

$\frac{d}{d x} [0.0000329 t^{3}] + \frac{d}{d x} [- 0.0061 t^{2}] + \frac{d}{d x} [0.0514 t] + \frac{d}{d x} [417]$ .

\frac{d}{d x} [0.0000329 t^{3}] + \frac{d}{d x} [- 0.0061 t^{2}] + \frac{d}{d x} [0.0514 t] + \frac{d}{d x} [417]

$\frac{d}{d x} [0.0000329 t^{3}] + \frac{d}{d x} [- 0.0061 t^{2}] + \frac{d}{d x} [0.0514 t] + \frac{d}{d x} [417]$

Step 1.1.2

Since

0.0000329 t^{3}

$0.0000329 t^{3}$ is constant with respect to

x

$x$ , the derivative of

0.0000329 t^{3}

$0.0000329 t^{3}$ with respect to

x

$x$ is

0

$0$ .

0 + \frac{d}{d x} [- 0.0061 t^{2}] + \frac{d}{d x} [0.0514 t] + \frac{d}{d x} [417]

$0 + \frac{d}{d x} [- 0.0061 t^{2}] + \frac{d}{d x} [0.0514 t] + \frac{d}{d x} [417]$

Step 1.1.3

Since

- 0.0061 t^{2}

$- 0.0061 t^{2}$ is constant with respect to

x

$x$ , the derivative of

- 0.0061 t^{2}

$- 0.0061 t^{2}$ with respect to

x

$x$ is

0

$0$ .

0 + 0 + \frac{d}{d x} [0.0514 t] + \frac{d}{d x} [417]

$0 + 0 + \frac{d}{d x} [0.0514 t] + \frac{d}{d x} [417]$

Step 1.1.4

Since

0.0514 t

$0.0514 t$ is constant with respect to

x

$x$ , the derivative of

0.0514 t

$0.0514 t$ with respect to

x

$x$ is

0

$0$ .

0 + 0 + 0 + \frac{d}{d x} [417]

$0 + 0 + 0 + \frac{d}{d x} [417]$

Step 1.1.5

Since

417

$417$ is constant with respect to

x

$x$ , the derivative of

417

$417$ with respect to

x

$x$ is

0

$0$ .

0 + 0 + 0 + 0

$0 + 0 + 0 + 0$

Step 1.1.6

Combine terms.

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Step 1.1.6.1

Add

0

$0$ and

0

$0$ .

0 + 0 + 0

$0 + 0 + 0$

Step 1.1.6.2

Add

0

$0$ and

0

$0$ .

0 + 0

$0 + 0$

Step 1.1.6.3

Add

0

$0$ and

0

$0$ .

f' (x) = 0

$f' (x) = 0$

f' (x) = 0

$f' (x) = 0$

f' (x) = 0

$f' (x) = 0$

Step 1.2

The first derivative of

f (x)

$f (x)$ with respect to

x

$x$ is

0

$0$ .

0

$0$

0

$0$

Step 2

Set the first derivative equal to

0

$0$ then solve the equation

0 = 0

$0 = 0$ .

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Step 2.1

Set the first derivative equal to

0

$0$ .

0 = 0

$0 = 0$

Step 2.2

Since

0 = 0

$0 = 0$ , the equation will always be true.

Always true

Step 3

There are no values of

x

$x$ in the domain of the original problem where the derivative is

0

$0$ or undefined.

No critical points found