Question

C(x)=x^3-2x^2+8x+50. find the critical numbers

Answer 1

$\large\begin{array}{l} \textsf{Finding the critical points of a given function:}\\\\ \mathsf{C(x)=x^3-2x^2+8x+50}\\\\\\ \bullet~~\textsf{1}\mathsf{^{st}}\textsf{ step: Find the derivative of C(x):}\\\\ \mathsf{C'(x)=(x^3-2x^2+8x+50)'}\\\\ \mathsf{C'(x)=3x^{3-1}-2\cdot 2x^{2-1}+8\cdot 1+0}\\\\ \mathsf{C'(x)=3x^2-4x+8} \end{array}$


$\large\begin{array}{l} \bullet~~\textsf{2}\mathsf{^{nd}}\textsf{ step: Solve the equation } \mathsf{C'(x)=0:}\\\\ \mathsf{3x^2-4x+8=0}\quad\Rightarrow\quad\begin{cases}\mathsf{a=3}\\\mathsf{b=-4}\\\mathsf{c=8} \end{cases}\\\\ \mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=(-4)^2-4\cdot 3\cdot 8}\\\\ \mathsf{\Delta=16-96}\\\\ \mathsf{\Delta=-80$


$\large\begin{array}{l} \textsf{Since the discriminant is a negative number, there's no}\\\textsf{real solution for }\mathsf{C'(x)=0.}\\\\\\ \textsf{Therefore, }\mathsf{C(x)}\textsf{ doesn't have any critical point.} \end{array}$


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Tags: critical points numbers function derivative differential calculus