{:[f^(')(x)=16x^(3)+6x-7" and "],[f(5)=2500.]:} f(-1)=

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{:[f^(')(x)=16x^(3)+6x-7" and "],[f(5)=2500.]:}  f(-1)=

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Integrate f'(x) to find f(x: To find f(-1), we need to evaluate the function f at x = -1. However, we are given f'(x), which is the derivative of f(x), not the function itself. Since we cannot directly find f(-1) from f'(x), we need to first find the general form of f(x) by integrating f'(x). Let's integrate f'(x) to find f(x).Find Constant C: The integral of f'(x) = 16x^3 + 6x - 7 is f(x) = (16/4)x^4 + (6/2)x^2 - 7x + C, where C is the constant of integration. Simplifying the integral, we get f(x) = 4x^4 + 3x^2 - 7x + C.Calculate C value: We are given that f(5) = 2500. We can use this information to find the value of the constant C. Plugging x = 5 into f(x), we get 2500 = 4(5)^4 + 3(5)^2 - 7(5) + C.Substitute x = -1: Calculating the values, we have 2500 = 4(625) + 3(25) - 35 + C. This simplifies to 2500 = 2500 + 75 - 35 + C.Evaluate f(-1): Now, we solve for C: 2500 = 2500 + 40 + C. Subtracting 2500 and 40 from both sides, we get C = 2500 - 2500 - 40.Calculating the value of C, we find that C = -40. Now we have the complete function f(x) = 4x^4 + 3x^2 - 7x - 40.Finally, we can find f(-1) by substituting x = -1 into the function f(x). f(-1) = 4(-1)^4 + 3(-1)^2 - 7(-1) - 40.Calculating the values, we get f(-1) = 4(1) + 3(1) + 7 - 40. This simplifies to f(-1) = 4 + 3 + 7 - 40.Adding and subtracting the numbers, we find that f(-1) = 14 - 40. f(-1) = -26.

$\begin{array}{l}f^{\prime}(x)=16 x^{3}+6 x-7 \text { and } f(5)=2500 .\end{array}$ $\newline$ $f(-1)=$

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f′(x)=16x3+6x−7 and f(5)=2500. \begin{array}{l}f^{\prime}(x)=16 x^{3}+6 x-7 \text { and } f(5)=2500 .\end{array} f′(x)=16x3+6x−7 and f(5)=2500.​\newlinef(−1)= f(-1)= f(−1)=

$\begin{array}{l}f^{\prime}(x)=16 x^{3}+6 x-7 \text { and } f(5)=2500 .\end{array}$ $\newline$ $f(-1)=$