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)

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) )

Accepted Answer

# Explanation:
## Step1: Differentiate the function
Using the power rule $(x^n)^\prime = nx^{n - 1}$, we have:
$$
\begin{align*}
f^\prime(x)&=\frac{1}{7}\times7x^{6}+\frac{1}{2}\times6x^{5}-5x^{4}-\frac{15}{4}\times4x^{3}+\frac{4}{3}\times3x^{2}+12x\
&=x^{6}+3x^{5}-5x^{4}-15x^{3}+4x^{2}+12x
\end{align*}
$$
## Step2: Calculate $f^\prime(-3.1)$
$$
\begin{align*}
f^\prime(-3.1)&=(-3.1)^{6}+3(-3.1)^{5}-5(-3.1)^{4}-15(-3.1)^{3}+4(-3.1)^{2}+12(-3.1)\
\end{align*}
$$
$$
\begin{align*}
(-3.1)^{6}&=(3.1)^{6}\approx887.503681\
3(-3.1)^{5}&=-3\times(3.1)^{5}\approx -3\times286.29151\approx - 858.87453\
-5(-3.1)^{4}&=-5\times(3.1)^{4}\approx-5\times92.3521\approx - 461.7605\
-15(-3.1)^{3}&=15\times(3.1)^{3}\approx15\times29.791\approx446.865\
4(-3.1)^{2}&=4\times9.61 = 38.44\
12(-3.1)&=-37.2
\end{align*}
$$
$$
\begin{align*}
f^\prime(-3.1)&\approx887.503681-858.87453 - 461.7605+446.865+38.44-37.2\
&\approx(887.503681 + 446.865+38.44)-(858.87453 + 461.7605+37.2)\
&\approx1372.808681 - 1357.83503\
&\approx14.973651
\end{align*}
$$
## Step3: Calculate $f^\prime(-1.5)$
$$
\begin{align*}
f^\prime(-1.5)&=(-1.5)^{6}+3(-1.5)^{5}-5(-1.5)^{4}-15(-1.5)^{3}+4(-1.5)^{2}+12(-1.5)\
\end{align*}
$$
$$
\begin{align*}
(-1.5)^{6}&=(1.5)^{6}=11.390625\
3(-1.5)^{5}&=-3\times(1.5)^{5}=-3\times7.59375=-22.78125\
-5(-1.5)^{4}&=-5\times(1.5)^{4}=-5\times5.0625=-25.3125\
-15(-1.5)^{3}&=15\times(1.5)^{3}=15\times3.375 = 50.625\
4(-1.5)^{2}&=4\times2.25 = 9\
12(-1.5)&=-18
\end{align*}
$$
$$
\begin{align*}
f^\prime(-1.5)&=11.390625-22.78125-25.3125 + 50.625+9-18\
&=(11.390625+50.625 + 9)-(22.78125+25.3125+18)\
&=71.015625 - 66.09375\
&=4.921875
\end{align*}
$$
## Step4: Calculate $f^\prime(0.4)$
$$
\begin{align*}
f^\prime(0.4)&=(0.4)^{6}+3(0.4)^{5}-5(0.4)^{4}-15(0.4)^{3}+4(0.4)^{2}+12(0.4)\
\end{align*}
$$
$$
\begin{align*}
(0.4)^{6}&=0.004096\
3(0.4)^{5}&=3\times0.01024 = 0.03072\
-5(0.4)^{4}&=-5\times0.0256=-0.128\
-15(0.4)^{3}&=-15\times0.064=-0.96\
4(0.4)^{2}&=4\times0.16 = 0.64\
12(0.4)&=4.8
\end{align*}
$$
$$
\begin{align*}
f^\prime(0.4)&=0.004096+0.03072-0.128-0.96 + 0.64+4.8\
&=(0.004096+0.03072+0.64+4.8)-(0.128 + 0.96)\
&=5.474816-1.088\
&=4.386816
\end{align*}
$$

# Answer:
Since $f^\prime(-3.1)\approx14.97$, $f^\prime(-1.5)\approx4.92$, $f^\prime(0.4)\approx4.39$, we have $f^\prime(0.4)<f^\prime(-1.5)<f^\prime(-3.1)$, so the answer is D.

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QUESTION IMAGE

Question

Explanation:

Step1: Differentiate the function

Step2: Calculate \(f^\prime(-3.1)\)

Step3: Calculate \(f^\prime(-1.5)\)

Step4: Calculate \(f^\prime(0.4)\)

Answer: