Question

find ( f^{prime}(0), f^{prime prime}(0) ), and determine whether ( f ) has a local minimum, local maximum, or neither at ( x = 0 ).

( f(x)=3 x^{5}-9 x^{2}+8 )

what is ( f^{prime}(0) ) ?

( f^{prime}(0)=0 )

what is ( f^{prime prime}(0) ) ?

( f^{prime prime}(0)=-18 )

does the function have a local minimum, a local maximum, or neither?

a. the function has a local minimum at ( x = 0 ).
b. the function has a local maximum at ( x = 0 ).
c. the function has neither a local minimum nor a local maximum at ( x = 0 ).

Explanation:

Step1: Find the first - derivative

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(y = f(x)=3x^{5}-9x^{2}+8\), we have \(f^\prime(x)=3\times5x^{4}-9\times2x+0 = 15x^{4}-18x\).
Substitute \(x = 0\) into \(f^\prime(x)\): \(f^\prime(0)=15\times0^{4}-18\times0=0\).

Step2: Find the second - derivative

Differentiate \(f^\prime(x)=15x^{4}-18x\) with respect to \(x\). Using the power rule again, \(f^{\prime\prime}(x)=15\times4x^{3}-18=60x^{3}-18\).
Substitute \(x = 0\) into \(f^{\prime\prime}(x)\): \(f^{\prime\prime}(0)=60\times0^{3}-18=-18\).

Step3: Use the second - derivative test

The second - derivative test states that if \(f^\prime(a) = 0\) and \(f^{\prime\prime}(a)>0\), then \(f(x)\) has a local minimum at \(x = a\); if \(f^\prime(a)=0\) and \(f^{\prime\prime}(a)<0\), then \(f(x)\) has a local maximum at \(x = a\); if \(f^\prime(a)=0\) and \(f^{\prime\prime}(a)=0\), the test is inconclusive.
Here, \(a = 0\), \(f^\prime(0) = 0\) and \(f^{\prime\prime}(0)=-18<0\).

Answer:

\(f^\prime(0)=0\), \(f^{\prime\prime}(0)=-18\), B. The function has a local maximum at \(x = 0\).