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)=square )

Explanation:

Step1: Find the first derivative

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

Step2: Find the second derivative

Differentiate \(f^\prime(x)=15x^{4}-18x\) again. Using the power rule, \(f^{\prime\prime}(x)=15\times4x^{3}-18\times1=60x^{3}-18\).

Step3: Evaluate the second derivative at \(x = 0\)

Substitute \(x = 0\) into \(f^{\prime\prime}(x)\). So \(f^{\prime\prime}(0)=60\times0^{3}-18=- 18\).

Answer:

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