Question #181117

Let 𝑓(𝑥) = 1 3 𝑥 3 + 𝑥 2 − 15𝑥 − 9 . Use detailed sign tables in answering the following questions. (a) Find the intervals in which 𝑓 is increasing or decreasing. (b) Find the intervals in which the graph of 𝑦 = 𝑓(𝑥) is concave upwards or downwards. 



Expert's answer

f(x)=13x3+x2−15x−9f(x)=13x^3+x^2-15x-9


a)

f′(x)=39x2+2x−15f'(x)=39x^2+2x-15


f′(x)=0f'(x)=0 when x=−2±4+4∗15∗392∗39=−1±58639x=\frac{-2\pm\sqrt{4+4*15*39}}{2*39}=\frac{-1\pm\sqrt{586}}{39}


f(x)f(x) is increasing when f′(x)>0f'(x)>0 or when

−∞<x<−1−58639-\infty<x<\frac{-1-\sqrt{586}}{39}

and when

−1+58639<x<+∞\frac{-1+\sqrt{586}}{39}<x<+\infty


f(x)f(x) is decreasing when f′(x)<0f'(x)<0 or when

−1−58639<x<−1+58639\frac{-1-\sqrt{586}}{39}<x<\frac{-1+\sqrt{586}}{39}


b)

f′′(x)=78x+2f''(x)=78x+2


f′′(x)=0f''(x)=0 when x=−139x=-\frac{1}{39}


When x<−139x<-\frac{1}{39} , f′′(x)<0f''(x)<0 , f(x)f(x) is concave upwards.


When x>−139x>-\frac{1}{39} , f′′(x)>0f''(x)>0 , f(x)f(x) is concave downwards.





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