let f be the function given by f(x)=1/7x^7 + 1/2x^6 - x^5 - 15/4x^4 + 4/3x^3 + 6x^2. which of the following…

Explanation:

Step1: Differentiate the function

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, if $f(x)=\frac{1}{7}x^{7}+\frac{1}{2}x^{6}-x^{5}-\frac{15}{4}x^{4}+\frac{4}{3}x^{3}+6x^{2}$, then $f'(x)=x^{6}+3x^{5}-5x^{4}-15x^{3}+4x^{2}+12x$.

Step2: Calculate $f'(-3.1)$

$f'(-3.1)=(-3.1)^{6}+3(-3.1)^{5}-5(-3.1)^{4}-15(-3.1)^{3}+4(-3.1)^{2}+12(-3.1)$ $=887.503681-837.97893 - 459.2405+438.915 + 38.44-37.2$ $=180.449251$.

Step3: Calculate $f'(-1.5)$

$f'(-1.5)=(-1.5)^{6}+3(-1.5)^{5}-5(-1.5)^{4}-15(-1.5)^{3}+4(-1.5)^{2}+12(-1.5)$ $=11.390625-15.1875 - 25.3125+50.625+9 - 18$ $=2.515625$.

Step4: Calculate $f'(0.4)$

$f'(0.4)=(0.4)^{6}+3(0.4)^{5}-5(0.4)^{4}-15(0.4)^{3}+4(0.4)^{2}+12(0.4)$ $=0.004096 + 0.03072-0.128 - 0.96+0.64+4.8$ $=3.376816$.

Step5: Compare the values

We have $f'(-1.5)=2.515625$, $f'(0.4)=3.376816$, $f'(-3.1)=180.449251$. So $f'(-1.5)<f'(0.4)<f'(-3.1)$.

Answer:

C. $f'(-1.5)<f'(0.4)<f'(-3.1)$