Answer:
6
Step-by-step explanation:
[tex]f(3-2x)=3x^2+1[/tex]
I'm going to replace [tex]3-2x[/tex] with [tex]u[/tex].
[tex]u=3-2x[/tex].
Solving this for [tex]x[/tex] gives:
[tex]\frac{u-3}{-2}=x[/tex] (I subtracted 3 on both sides and then divided both sides by -2).
Writing the function in terms of [tex]u[/tex] gives:
[tex]f(u)=3(\frac{u-3}{-2})^2+1[/tex]
Now let's differentiate.
[tex]f'(u)=3 \cdot 2(\frac{u-3}{-2})^{2-1} \cdot \frac{1}{-2}+0[/tex] by chain rule and constant rule.
[tex]f'(u)=6(\frac{u-3}{-2})\cdot \frac{1}{-2}[/tex]
[tex]f'(u)=-3(u-3)\cdot \frac{1}{-2}[/tex]
[tex]f'(u)=\frac{3}{2}(u-3)[/tex]
Now let's evaluate our function for [tex]u=7[/tex]
[tex]f'(7)=\frac{3}{2}(7-3)[/tex]
[tex]f'(7)=\frac{3}{2}(4)[/tex]
[tex]f'(7)=6[/tex]
Or an easier solution:
[tex]f(3-2x)=3x^2+1[/tex]
[tex]-2 \cdot f'(3-2x)=6x[/tex]
I applied chain rule on the left and power rule on the right.
Now [tex]3-2x[/tex] is 7 when [tex]x=-2[/tex].
[tex]-2 \cdot f'(3-2(-2))=6(-2)[/tex]
[tex]-2 \cdot f'(3+4)=6(-2)[/tex]
[tex]-2 \cdot f'(7)=-12[/tex]
[tex]f'(7)=6[/tex]
