Respuesta :
Solution
- The critical points of a graph are the points where the turning points of the graph are zero.
- The turning points are gotten by differentiating the function after which we can equate to zero.
- This is done below:
[tex]f(x)=2x^3-30x^2+126x-10[/tex]- The way to differentiate is given below:
[tex]\begin{gathered} Given\text{ }f(x)=x^n \\ f^{\prime}(x)=nx^{n-1} \\ \\ \text{ Thus, to differentiate the function }ax^n+bx^{n-1}+cx^{n-2}+dx^{n-3}... \\ f^{\prime}(x)=nax^{n-1}+b(n-1)x^{n-2}+c(n-2)x^{n-3}+... \end{gathered}[/tex]For example,
[tex]\begin{gathered} f(x)=5x+6 \\ \text{ Differentiating, we have:} \\ f^{\prime}(x)=5(1)x^{1-1}=5 \\ \\ f(x)=10x^2+3x-20 \\ \text{ Differentiating, we have:} \\ f^{\prime}(x)=10(2)x^{2-1}+3(1)x^{1-1} \\ f^{\prime}(x)=20x+3 \end{gathered}[/tex]- The differentiation is done below:
[tex]f^{\prime}(x)=6x^2-60x+126[/tex]- The critical points are where f'(x)= 0. Thus, we have:
[tex]\begin{gathered} 6x^2-60x+126=0 \\ 6x^2-42x-18x+126=0 \\ 6x(x-7)-18(x-7)=0 \\ (x-7)(6x-18)=0 \\ \\ x=7\text{ or }6x=18 \\ x=7\text{ or }x=3 \end{gathered}[/tex]Final Answer
The critical points are x = 7 and x = 3
