1 Answer

HenryM · Answer 1 · 2023-05-27T09:59:59+0000

The function given is:

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:

$Maximum:f^(\prime)(1)=0$

Since we know that the point (1, 6) lies in the graph of the function, we can write:

$\begin{gathered} 6=a\cdot1^3+b\cdot1 \\ . \\ 6=a+b \end{gathered}$

We have an equation that relates a and b, we need to find another one. Let's differentiate f(x):

$(d)/(dx)f(x)=(d)/(dx)(ax^3+bx)=(d)/(dx)(ax^3)+(d)/(dx)(bx)=a\cdot(d)/(dx)(x^3)+b\cdot(d)/(dx)x=a\cdot3x^(3-1)+b\cdot1x^(1-1)=3ax^2+b$

Thus:

$f^(\prime)(x)=3ax^2+b$

We know that f'(1) = 0, thus:

$\begin{gathered} 0=3a\cdot1^2+b \\ . \\ 0=3a+b \end{gathered}$

Now, we have a system of two linear equations:

$\begin{cases}6={a+b} \\ 0={3a+b}\end{cases}$

We can subtract the second equation to the first one, to solve by elimination:

And solve:

$\begin{gathered} 6=a-3a+b-b \\ . \\ 6=-2a \\ . \\ a=(6)/(-2)=-3 \end{gathered}$

We have found the value of a = -3. Now, we can find b, using the first equation:

$\begin{gathered} 6=-3+b \\ . \\ b=6+3=9 \end{gathered}$

Thus, the answer is:

a = -3

b = 9

We can also check if the result we get is correct:

$\begin{cases}a={-3} \\ b={9}\end{cases}\Rightarrow\begin{cases}f(x)={-3x^3}+9x \\ f^(\prime)(x)={-9x^2+9}\end{cases}$

And we should get f(1) = 6 and f'(1) = 0:

$\begin{gathered} f(1)=-3\cdot1^3+9\cdot1=-3+9=6 \\ f^(\prime)(1)=-9\cdot1+9=-9+9=0 \end{gathered}$

The values we got are correct.

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