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Given: $f(x)=2 x^{3}-2 x^{2}+4 x-1$. Determine the interval on which $f$ is concave Up.
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\begin{array}{l}
\begin{array}{l}
f(x)=2 x^{3}-2 x^{2}+4 x-1 \\
f^{\prime}(x)=6 x^{2}-4 x+4 \\
f^{\prime \prime}(x)=12 x-4
\end{array}\\
f \text { is concaveup when } f^{\prime \prime}(x)>0\\
\therefore 12 x-4>0\\
\begin{array}{r}
12 x>4 \\
x>\frac{1}{3}
\end{array}
\end{array}

Explanation:

This question requires in-depth understanding of the concept of concavity. You need to have both a visual and conceptual understanding of what it means to say that a function is concave up in a given interval or concave down in a given interval. To answer this question, you need to be conversant with the logical procedure of determining the interval on which a function is concave up. This higher order reasoning process entails the following procedural steps: You must first determine $f^{\prime}(x)$, and then calculate $f^{\prime \prime}(x)$, and finally solve $f^{\prime \prime}(x)>0$.

by Diamond (75,948 points)

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