Question

let ( f ) be the function given by ( f(x)=\frac{1}{7} x^{7}+\frac{1}{2} x^{6}-x^{5}-\frac{15}{4} x^{4}+\frac{4}{3} x^{3}+6 x^{2} ). which of the following statements is true?
a ( f^{prime}(-3.1)<f^{prime}(-1.5)<f^{prime}(0.4) )
b ( f^{prime}(-3.1)<f^{prime}(0.4)<f^{prime}(-1.5) )
c ( f^{prime}(-1.5)<f^{prime}(0.4)<f^{prime}(-3.1) )
d ( f^{prime}(0.4)<f^{prime}(-1.5)<f^{prime}(-3.1) )

Explanation:

Step1: Find the derivative of \( f(x) \)

Given \( f(x)=\frac{1}{7}x^{7}+\frac{1}{2}x^{6}-x^{5}-\frac{15}{4}x^{4}+\frac{4}{3}x^{3}+6x^{2} \).
Using the power rule \( (x^{n})^\prime = nx^{n - 1} \), we have \( f^\prime(x)=x^{6}+3x^{5}-5x^{4}-15x^{3}+4x^{2}+12x \).
Factor \( f^\prime(x)=x(x^{5}+3x^{4}-5x^{3}-15x^{2}+4x + 12)=x(x + 3)(x^{4}-5x^{2}+4)=x(x + 3)(x^{2}-1)(x^{2}-4)=x(x + 3)(x - 1)(x + 1)(x - 2)(x + 2) \).

Step2: Analyze the sign of \( f^\prime(x) \)

We can use a sign - chart or test values in the intervals separated by the critical points \( x=-3,-2,-1,0,1,2 \).
Let's evaluate \( f^\prime(-3.1)=(-3.1)(-3.1 + 3)(-3.1-1)(-3.1 + 1)(-3.1-2)(-3.1 + 2)\).
\(f^\prime(-3.1)=(-3.1)(-0.1)(-4.1)(-2.1)(-5.1)(-1.1)\lt0\).
Evaluate \( f^\prime(-1.5)=(-1.5)(-1.5 + 3)(-1.5-1)(-1.5 + 1)(-1.5-2)(-1.5 + 2)\).
\(f^\prime(-1.5)=(-1.5)(1.5)(-2.5)(-0.5)(-3.5)(0.5)\gt0\).
Evaluate \( f^\prime(0.4)=(0.4)(0.4 + 3)(0.4-1)(0.4 + 1)(0.4-2)(0.4 + 2)\).
\(f^\prime(0.4)=(0.4)(3.4)(-0.6)(1.4)(-1.6)(2.4)\gt0\).
Since \( y = f^\prime(x) \) is a polynomial, and we know that for \( x\lt - 3\), \( f^\prime(x)\lt0\); for \( - 3\lt x\lt - 2\), \( f^\prime(x)\gt0\); for \( - 2\lt x\lt - 1\), \( f^\prime(x)\lt0\); for \( - 1\lt x\lt0\), \( f^\prime(x)\gt0\); for \( 0\lt x\lt1\), \( f^\prime(x)\lt0\); for \( 1\lt x\lt2\), \( f^\prime(x)\gt0\); for \( x\gt2\), \( f^\prime(x)\gt0\).
We can also use the fact that \( f^\prime(x) \) is a continuous function.
Another way is to calculate the values:
\(f^\prime(-3.1)=(-3.1)^{6}+3(-3.1)^{5}-5(-3.1)^{4}-15(-3.1)^{3}+4(-3.1)^{2}+12(-3.1)\).
\(f^\prime(-3.1)=887.503681-3\times241.8651-5\times92.3521 + 15\times29.791+4\times9.61-37.2\).
\(f^\prime(-3.1)=887.503681 - 725.5953-461.7605 + 446.865+38.44-37.2\lt0\).
\(f^\prime(-1.5)=(-1.5)^{6}+3(-1.5)^{5}-5(-1.5)^{4}-15(-1.5)^{3}+4(-1.5)^{2}+12(-1.5)\).
\(f^\prime(-1.5)=11.390625-3\times7.59375-5\times5.0625 + 15\times3.375+4\times2.25-18\).
\(f^\prime(-1.5)=11.390625-22.78125 - 25.3125+50.625+9 - 18\gt0\).
\(f^\prime(0.4)=(0.4)^{6}+3(0.4)^{5}-5(0.4)^{4}-15(0.4)^{3}+4(0.4)^{2}+12(0.4)\).
\(f^\prime(0.4)=0.004096+3\times0.01024-5\times0.0256-15\times0.064 + 4\times0.16+4.8\).
\(f^\prime(0.4)=0.004096 + 0.03072-0.128-0.96+0.64+4.8\gt0\).
Since \( f^\prime(-3.1)\lt0\), \( f^\prime(-1.5)\gt0\), \( f^\prime(0.4)\gt0\), and \( y = f^\prime(x) \) is a continuous function, we have \( f^\prime(-3.1)\lt f^\prime(0.4)\lt f^\prime(-1.5)\).

Answer:

B. \( f^\prime(-3.1)\lt f^\prime(0.4)\lt f^\prime(-1.5)\)