Question

find the x and y coordinates of all inflection points.

$f(x)=2x^{\frac{9}{5}} + 2$

what is/are the inflection point(s)? select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. the inflection point(s) is/are
(type an ordered pair. use a comma to separate answers as needed.)
b. there are no inflection points.

Explanation:

Step1: Find the first derivative

Use the power rule \( (x^n)^\prime=nx^{n - 1}\). For \(y = 2x^{\frac{9}{5}}+2\), the first derivative \(y^\prime=2\times\frac{9}{5}x^{\frac{9}{5}-1}=\frac{18}{5}x^{\frac{4}{5}}\).

Step2: Find the second derivative

Differentiate \(y^\prime=\frac{18}{5}x^{\frac{4}{5}}\) again using the power rule. \(y^{\prime\prime}=\frac{18}{5}\times\frac{4}{5}x^{\frac{4}{5}-1}=\frac{72}{25}x^{-\frac{1}{5}}=\frac{72}{25x^{\frac{1}{5}}}\).

Step3: Analyze the second - derivative for inflection points

An inflection point occurs where \(y^{\prime\prime} = 0\) or \(y^{\prime\prime}\) is undefined.
Set \(y^{\prime\prime}=0\), \(\frac{72}{25x^{\frac{1}{5}}}=0\). There is no solution for \(x\) since the numerator \(72
eq0\).
The second derivative \(y^{\prime\prime}=\frac{72}{25x^{\frac{1}{5}}}\) is undefined at \(x = 0\). But we need to check the concavity on either side of \(x = 0\).
For \(x>0\), say \(x = 1\), \(y^{\prime\prime}(1)=\frac{72}{25}>0\) (function is concave up).
For \(x<0\), say \(x=-1\), \(y^{\prime\prime}(-1)=\frac{72}{25(-1)^{\frac{1}{5}}}=-\frac{72}{25}<0\) (function is concave down). But \(x = 0\) is not in the domain of \(y^{\prime\prime}\) in the context of real - valued functions for the second - derivative (since \(x^{-\frac{1}{5}}=\frac{1}{\sqrt[5]{x}}\) and \(x = 0\) makes it undefined, and also for the original function \(y = 2x^{\frac{9}{5}}+2=2\sqrt[5]{x^{9}}+2\), the second - derivative change at \(x = 0\) does not correspond to a point on the function in the sense of the standard definition of inflection points (where the function is differentiable at that point).

Answer:

B. There are no inflection points.