Question

Determine the intervals on which the function f(x)=−7x2−7x+6f(x)=−7x2−7x+6 is concave up or down and find the points of inflection. (Use symbolic notation and fractions where needed. Give your answer as comma separated list. Enter NULL if there are no points of inflection.) Points of inflection: help (fractions) (Use symbolic notation and fractions where needed. Give your answers as intervals in the form (*,*). Use inf for infinity , U for combining intervals, and appropriate type of parenthesis "(", ")", "[" or "]" depending on whether the interval is open or closed. Enter NULL if interval is empty.) f(x)f(x) is concave up when x∈x∈ help (intervals) f(x)f(x) is concave down when x∈

Answer 1

• Point of Inflection: Does not Exist
• f(x) is concave up when x∈(-∞,-6)
• f(x) is concave down when x∈(-6, ∞)

Explanation

We are to determine the intervals on which the function [TeX]f(x)=\frac{-7x^2-7}{x+6}[/TeX] is concave up or down and find the points of inflection.

A function f(x) is concave up in the intervals for which the second derivative of the function [TeX]\frac{d^{2}}{dx^{2}} > 0 [/TeX].

A function f(x) is concave down in the intervals for which the second derivative of the function [TeX]\frac{d^{2}}{dx^{2}} < 0 [/TeX].

Given [TeX]f(x)=\frac{-7x^2-7}{x+6}[/TeX]

We need to solve for the second derivative of the function.

The first derivative:

Using Quotient Rule:

[TeX]f^{'}(x)=\frac{14x(x+6)-(7x^2-7)}{(x+6)^{2}} [/TeX].

[TeX]f^{'}(x)=\frac{7x^2+84x+7}{(x+6)^{2}} [/TeX].

The Second derivative:

Similarly applying quotient rule:

[TeX]f^{''}(x )=\frac{(x+6)^{2}(14x+84)-( 7x^2+84x+7)(2x+12)}{(x+6)^{4}} [/TeX]

[TeX]f^{''}(x)=\frac{14(x+6) (x+6) (x+6) -14(x+6)(x^2+12x+1) }{(x+6)^{4}} [/TeX]

[TeX]f^{''}(x )=\frac{14(x+6) [(x+6) (x+6) -x^2+12x+1)] }{(x+6)^{4}} [/TeX]

[TeX]f^{''}(x)=\frac{490 }{(x+6)^{3}} [/TeX]

Now [TeX]f^{''}(x)=\frac{490 }{(x+6)^{3}} =0[/TeX] is 490, thus [TeX]f^{''}(x) =0[/TeX] does not have a solution,

However, we determine the point where [TeX]f^{''}(x) [/TeX] is undefined. This point is x=-6.

In the Interval, (-∞,-6), f(x) is concave Up and in the Interval (-6, ∞), f(x) is concave down. It has no Inflection point.