6. Graphical Applications of Derivatives

The First Derivative Test

6. Graphical Applications of Derivatives

The First Derivative Test

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Identify the open intervals on which the function is increasing or decreasing.
$f(x)=x^{2/3}(4-x)$

Decreasing on $$\left$(-$\infty$,$\infty$$\right$)$

Increasing on $$\left$(-$\infty$,$\infty$$\right$)$

Increasing on $$\left$(-$\infty$,0$\right$)\&$\left$($\frac$85,$\infty$$\right$)$ , Decreasing on $$\left$(0,$\frac$85$\right$)$

Increasing on $$\left$(0,$\frac$85$\right$)$ , Decreasing on $$\left$(-$\infty$,0$\right$)\&$\left$($\frac$85,$\infty$$\right$)$

Guida verificata passo dopo passo

Step 1: Start by finding the derivative of the function f(x) = x^(2/3)(4 - x). Use the product rule for derivatives, which states that if f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x). Here, u(x) = x^(2/3) and v(x) = (4 - x).

Step 2: Compute the derivative of u(x) = x^(2/3). Using the power rule, the derivative is u'(x) = (2/3)x^(-1/3). Next, compute the derivative of v(x) = (4 - x), which is v'(x) = -1.

Step 3: Substitute u'(x), u(x), v'(x), and v(x) into the product rule formula. This gives f'(x) = [(2/3)x^(-1/3)(4 - x)] + [x^(2/3)(-1)]. Simplify this expression to get the derivative in a usable form.

Step 4: Set f'(x) = 0 to find the critical points. Solve the equation [(2/3)x^(-1/3)(4 - x)] - x^(2/3) = 0. Factorize the expression to isolate x and determine the critical points.

Step 5: Use the critical points to test the intervals on the number line. Determine the sign of f'(x) in each interval to identify where the function is increasing (f'(x) > 0) or decreasing (f'(x) < 0). Based on this analysis, conclude the intervals of increase and decrease for the function.