[tex] \text{Consider the function}\\ \\ f(x)=2x^3-30x^2+126x+2\\ \\ \text{To find the local maxima, first we find the critical points.}\\ \text{for critical points set the first derivative 0 and solve for x.}\\ \text{so using power rule of derivatives, we have}\\ \\ f'(x)=2(3x^2)-30(2x)+126(1)+0 [/tex]
[tex] \Rightarrow f'(x)=6x^2-60x+126\\ \\ \text{now for critical points, we set }f'(x)=0\\ \\ \Rightarrow 6x^2-60x+126=0\\ \\ \Rightarrow 6(x^2-10x+21)=0\\ \\ \Rightarrow x^2-10x+21=0\\ \\ \Rightarrow x^2-7x-3x+21=0\\ \\ \Rightarrow x(x-7)-3(x-7)=0\\ \\ \Rightarrow (x-3)(x-7)=0 [/tex]
[tex] \Rightarrow x-3=0, \ \ \ \ \ \ \ \ \text{ and } \ \ \ \ \ x-7=0\\ \\ \Rightarrow x=3, \ \ \ \ \ \ \ \ \text{ and } \ \ \ \ \ x=7\\ \\ \text{now to check for maximum minimum, we do second derivative test.}\\ \\ f''(x)=12x-60\\ \\ \text{at x=3, second derivative is}\\ \\ f''(3)=12(3)-60=-24 <0, \text{ so local maximum occur at x=3}\\ \\ \text{and at x=7, second derivative is ,} [/tex]
[tex] f''(7)=12(7)-60=24>0, \text{ so local minimum occur at x=7.}\\ \\ \text{hence the local maximum value of the function is}\\ \\ f(3)=2(3)^3-30(3)^2+126(3)+2=164 [/tex]
Hence the local maxima of f is: 164
