3.4 Lösning 6b

$$\begin{array}{rcl} f(x) & = & x^4\, (1 \, - \, x) \, = \, x^4 \, - \, x^5 \\ f'(x) & = & 4\,x^3 \, - \, 5\,x^4 \\ \end{array}$$

$$\begin{array}{rcl} 4\,x^3 \, - \, 5\,x^4 & = & 0 \\ x^3 \, (4 \,- \, 5\,x) & = & 0 \\ x_1 & = & 0 \\ 4 \,- \, 5\,x_2 & = & 0 \\ 4 & = & 5\,x_2 \\ 4 / 5 & = & x_2 \\ x_2 & = & 0,8 \end{array}$$

$$\Rightarrow \; f(x) \; {\rm har\;en\;kritisk\;punkt\;till:} \;\; x_2 = 0,8 \; {\rm .}$$

$$f''(x) \, = \, 12\,x^2 - 20\,x^3$$

$$f''(0,8) \, = \, 12 \cdot 0,8^2 - 20 \cdot 0,8^3 = -2,56 < 0$$

$$\Rightarrow \quad x_2 = 0,8 \quad {\rm är\;en\;maximipunkt} {\rm .}$$ .