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Sagot :
Answer:
[tex]f(x)=\frac{1}{2}(2x-9)^4+5[/tex]Explanation:
Given f'(x) defined below:
[tex]f^{\prime}(x)=4(2x-9)^3_{}[/tex]First, integrate f'(x) to find f(x).
[tex]\int f^{\prime}(x)=\int 4(2x-9)^3dx=4\int (2x-9)^3dx[/tex]Let u = 2x-9
[tex]u=2x-9\implies du=2dx\implies dx=\frac{du}{2}[/tex]Thus:
[tex]\begin{gathered} f(u)=4\int u^3\frac{du}{2}=\frac{4}{2}\int u^3du=\frac{2u^4}{4}=\frac{1}{2}u^4+C \\ \implies f(u)=\frac{1}{2}u^4+C \end{gathered}[/tex]Replace u=2x-9.
[tex]f(x)=\frac{1}{2}(2x-9)^4+C[/tex]Next, using the point (5,11/2), we find the value of C, the constant of integration.
At (5, 11/2)
[tex]\begin{gathered} x=5,f(x)=\frac{11}{2} \\ f(x)=\frac{1}{2}(2x-9)^4+C \\ \frac{11}{2}=\frac{1}{2}(2\lbrack5\rbrack-9)^4+C \\ \frac{11}{2}=\frac{1}{2}(10-9)^4+C \\ \frac{11}{2}=\frac{1}{2}(1)^4+C \\ \frac{11}{2}=\frac{1}{2}+C \\ C=\frac{11}{2}-\frac{1}{2}=\frac{10}{2}=5 \end{gathered}[/tex]Therefore, the function f(x) is:
[tex]f(x)=\frac{1}{2}(2x-9)^4+5[/tex]
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