Calculus Examples

$f (x) = (4 - t^{2}) (1 + 5 t^{2})$

Step 1

Find the first derivative.

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Step 1.1

Since $(4 - t^{2}) (1 + 5 t^{2})$ is constant with respect to $x$ , the derivative of $(4 - t^{2}) (1 + 5 t^{2})$ with respect to $x$ is $0$ .

$f' (x) = 0$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $0$ .

$0$

Step 2

Set the first derivative equal to $0$ then solve the equation $0 = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$0 = 0$

Step 2.2

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

Always true

Step 3

There are no values of $x$ in the domain of the original problem where the derivative is $0$ or undefined.

No critical points found

Step 4

No points make the derivative $f' (x) = 0$ equal to $0$ or undefined. The interval to check if $f (x) = (4 - t^{2}) (1 + 5 t^{2})$ is increasing or decreasing is $(- \infty, \infty)$ .

$(- \infty, \infty)$

Step 5

Substitute any number, such as $1$ , from the interval $(- \infty, \infty)$ in the derivative $f' (x) = 0$ to check if the result is negative or positive. If the result is negative, the graph is decreasing on the interval $(- \infty, \infty)$ . If the result is positive, the graph is increasing on the interval $(- \infty, \infty)$ .

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Step 5.1

Replace the variable $x$ with $1$ in the expression.

$f' (1) = 0$

Step 5.2

The final answer is $0$ .

$0$

Step 6

The result of substituting $1$ into $f' (x) = 0$ is $0$ , which is positive, so the graph is increasing on the interval $(- \infty, \infty)$ .

Increasing on $(- \infty, \infty)$ since $0 > 0$

Step 7

Increasing over the interval $(- \infty, \infty)$ means that the function is always increasing.

Always Increasing

Step 8