The derivative of a function $f$ is given by $f^{\prime}(x)=x^{4}-x^{3}-2 x^{2}+4 x-2$.

Where does the graph of $f$ have a relative minimum? Use a graphing calculator.

Choose 1 answer:
(A) $x=-2$
(B) $x=-1.769$
(C) $x=-1.067$
(D) $x=1$

Question

The derivative of a function $f$ is given by $f^{\prime}(x)=x^{4}-x^{3}-2 x^{2}+4 x-2$.

Where does the graph of $f$ have a relative minimum? Use a graphing calculator.

Choose 1 answer:
(A) $x=-2$
(B) $x=-1.769$
(C) $x=-1.067$
(D) $x=1$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the relative minimum of the function $f$, we need to analyze the critical points of its derivative $f'(x)$. ‖ ‖ step 2 ⋮ The derivative $f'(x)$ is given by $f'(x) = x^4 - x^3 - 2x^2 + 4x - 2$. ‖ ‖ step 3 ⋮ According to the asksia-ll calculator, the function $f$ has a relative minimum at $x \approx -1.06659$. ‖ ‖ step 4 ⋮ Comparing this result with the given options, the closest value to $x \approx -1.06659$ is (C) $x = -1.067$. ‖ ***C*** ∻Key Concept∻ ⚹Relative Minimum⚹ ∻Explanation∻ ⚹The relative minimum of a function occurs at a critical point where the derivative changes from negative to positive, indicating a trough in the graph of the function.⚹◊What method can be used to find the x-value where a function has a relative minimum given its derivative?⍭ Generate me a similar question◊

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