Question

1. no calculator the function f given by f(x)=3x^{5}-4x^{3}-3x has a relative maximum at x=

Explanation:

Step1: Find the derivative of the function

Using the power rule \( (x^n)^\prime=nx^{n - 1} \), for \( f(x)=3x^{5}-4x^{3}-3x \), we have \( f^\prime(x)=15x^{4}-12x^{2}-3 \).
Factor out a \( 3 \): \( f^\prime(x)=3(5x^{4}-4x^{2}-1) \).
Let \( t = x^{2} \), then the equation becomes \( 5t^{2}-4t - 1=0 \).
Using the quadratic formula \( t=\frac{4\pm\sqrt{16+20}}{10}=\frac{4\pm\sqrt{36}}{10}=\frac{4\pm6}{10} \).
We get \( t_1 = 1,t_2=-\frac{1}{5} \). Since \( t=x^{2}\geq0 \), when \( t = 1 \), \( x^{2}=1\), so \( x=\pm1 \).

Step2: Use the second - derivative test

Find the second - derivative \( f^{\prime\prime}(x)=60x^{3}-24x \).
When \( x = 1 \), \( f^{\prime\prime}(1)=60\times1^{3}-24\times1=36>0 \), so \( x = 1 \) is a relative minimum.
When \( x=-1 \), \( f^{\prime\prime}(-1)=60\times(-1)^{3}-24\times(-1)=-60 + 24=-36<0 \).

Answer:

\( x=-1 \)