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Sagot :
The function given is:
[tex]f(x)=ax^3+bx[/tex]And has a local maximum at (1, 6). Since it has a local extrema in x = 1, this means that the derivative of f(x) at x = 1 is 0:
[tex]Maximum:f^{\prime}(1)=0[/tex]Since we know that the point (1, 6) lies in the graph of the function, we can write:
[tex]\begin{gathered} 6=a\cdot1^3+b\cdot1 \\ . \\ 6=a+b \end{gathered}[/tex]We have an equation that relates a and b, we need to find another one. Let's differentiate f(x):
[tex]\frac{d}{dx}f(x)=\frac{d}{dx}(ax^3+bx)=\frac{d}{dx}(ax^3)+\frac{d}{dx}(bx)=a\cdot\frac{d}{dx}(x^3)+b\cdot\frac{d}{dx}x=a\cdot3x^{3-1}+b\cdot1x^{1-1}=3ax^2+b[/tex]Thus:
[tex]f^{\prime}(x)=3ax^2+b[/tex]We know that f'(1) = 0, thus:
[tex]\begin{gathered} 0=3a\cdot1^2+b \\ . \\ 0=3a+b \end{gathered}[/tex]Now, we have a system of two linear equations:
[tex]\begin{cases}6={a+b} \\ 0={3a+b}\end{cases}[/tex]We can subtract the second equation to the first one, to solve by elimination:
[tex]6-0=a+b-(3a+b)[/tex]And solve:
[tex]\begin{gathered} 6=a-3a+b-b \\ . \\ 6=-2a \\ . \\ a=\frac{6}{-2}=-3 \end{gathered}[/tex]We have found the value of a = -3. Now, we can find b, using the first equation:
[tex]\begin{gathered} 6=-3+b \\ . \\ b=6+3=9 \end{gathered}[/tex]Thus, the answer is:
a = -3
b = 9
We can also check if the result we get is correct:
[tex]\begin{cases}a={-3} \\ b={9}\end{cases}\Rightarrow\begin{cases}f(x)={-3x^3}+9x \\ f^{\prime}(x)={-9x^2+9}\end{cases}[/tex]And we should get f(1) = 6 and f'(1) = 0:
[tex]\begin{gathered} f(1)=-3\cdot1^3+9\cdot1=-3+9=6 \\ f^{\prime}(1)=-9\cdot1+9=-9+9=0 \end{gathered}[/tex]The values we got are correct.
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