Question

If $f\left(x\right)=a(x^2-1)(ax+b),a\ne 0$, has local maximum and minimum at $x=1$ and $x=2$ respectively, then $f\text{'}\left(x\right)$ is:

Answer 1

Step 1: Solve for $f\text{'}\left(x\right)$

Given: $f\left(x\right)=a(x^2-1)(ax+b),a\ne 0$ has a local maximum and minimum at $x_1$ and $x_2$.

$$\begin{gathered} f\left(x\right)=a(x^2-1)(ax+b) \\ \frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left[a(x^2-1)(ax+b)\right] \\ f\left(x\right)\text{'}=a\frac{d}{dx}\left[(x^2-1)(ax+b)\right] \end{gathered}$$

applying the product rule of differentiation

$$\begin{gathered} f\text{'}\left(x\right)=a\left[(ax+b)\frac{d}{dx}(x^2-1)+(x^2-1)\frac{d}{dx}(ax+b)\right] \\ f\text{'}\left(x\right)=a\left[(ax+b)·2x+(x^2-1)a\right] \\ f\text{'}\left(x\right)=a\left[2a{x}^2+2bx+a{x}^2-a\right] \\ f\text{'}\left(x\right)=a\left[3a{x}^2+2bx-a\right] \end{gathered}$$

Step 2: Solve for $f\text{'}\left(1\right)$

putting $x=1$

$$\begin{gathered} f\text{'}\left(1\right)=a\left[3a\times {\left(1\right)}^2+2b\left(1\right)-a\right] \\ f\text{'}\left(1\right)=a\left[3a+2b-a\right] \\ f\text{'}\left(1\right)=2a(a+b) \end{gathered}$$

Step 3: Solve for $f\text{'}\left(2\right)$

putting $x=2$

$$\begin{gathered} f\text{'}\left(2\right)=a\left[3a\times {\left(2\right)}^2+2b\left(2\right)-a\right] \\ f\text{'}\left(2\right)=a\left[12a+4b-a\right] \\ f\text{'}\left(2\right)=a(11a-4b) \end{gathered}$$

Hence,$f\text{'}\left(x\right)$ for $x=1,2$ is $2a(a+b)$ and $a(11a-4b)$.