`f(x) = xsqrt(x + 3)` Find the points of inflection and discuss the concavity of the graph of the function.

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embizze | High School Teacher | (Level 2) Educator Emeritus

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Find the inflection points and discuss the concavity for the function `f(x)=xsqrt(x+3) `

Inflection points occur where the second derivative is zero (and changes sign.)

Rewrite as ` f(x)=x(x+3)^(1/2) `

Using the product rule we get:

`f'(x)=(x+3)^(1/2)+x(1/2)(x+3)^(-1/2) `

`f''(x)=1/2(x+3)^(-1/2)+1/2(x+3)^(-1/2)+x/2(-1/2)(x+3)^(-3/2) `

`=(x+3)^(-1/2)-x/4(x-3)^(-3/2) `

`=(x+3)^(-3/2)(x+3-x/4) `

Setting f''(x)=0 we get:

x=-3 or x=-4. Since x=-4 is not in the domain the only zero is x=-3. This is the endpoint of the domain, so there are no inflection points. The second derivative is positive on the functions domain, so the function is concave up everywhere.

The graph: