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Find the second derivative of the function.\newlinef(x)=x3f(x) = x^3\newlinef′′(x)=f''(x) = ______

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Q. Find the second derivative of the function.\newlinef(x)=x3f(x) = x^3\newlinef′′(x)=f''(x) = ______
  1. Identify total tape needed: Step 11: Identify the total amount of tape needed and the amount per roll.\newlineTotal tape needed = 8,000 cm8,000 \, \text{cm}, Tape per roll = 2,000 cm2,000 \, \text{cm}.\newlineWe need to divide the total tape needed by the amount in each roll to find out how many rolls are required.\newlineCalculation: 8,000÷2,000=48,000 \div 2,000 = 4.

More problems from Find second derivatives of trigonometric, exponential, and logarithmic functions

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f′(x)=12ex and f(4)=−16+12e4.f(0)= \begin{array}{l}f^{\prime}(x)=12 e^{x} \text { and } f(4)=-16+12 e^{4} . \\ f(0)=\end{array}
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Solve for nn.125=n8\frac{12}{5}=\frac{n}{8} \newline n=□n = \Box
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The second derivative of the function f f is defined by f′′(x)=x2−5cos⁡(2x) f^{\prime \prime}(x)=x^{2}-5 \cos (2 x) for −2.5<x<3.5 -2.5<x<3.5 . Find the x x -values, if any, in the given domain where the function f f has an inflection point. You may use a calculator and round all values to 33 decimal places.\newlineAnswer: x= x=
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f′(x)=12ex and f(4)=−16+12e4.f(0)= \begin{array}{l}f^{\prime}(x)=12 e^{x} \text { and } f(4)=-16+12 e^{4} . \\ f(0)=\end{array}
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h(x)=x2+xg(x)=3x+5 Find (h⋅g)(x)\begin{array}{l}h(x)=x^{2}+x \\ g(x)=3 x+5 \\ \text{ Find }(h \cdot g)(x)\end{array}
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Factor completely:\newline8c3+27m6 8 c^{3}+27 m^{6} \newlineAnswer:
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