Answer:
[tex]f'(-1) = \pm 2\sqrt{2}[/tex]
Step-by-step explanation:
For [tex]f: \mathbb{R} \rightarrow \mathbb{R}[/tex] such that [tex]f(x) = 7-x^2[/tex], find [tex]f'(-1)[/tex]
Recall that
[tex]f'(y)=x \text{ whenever } f(x) = y, \forall y \in \mathbb{R}[/tex]
Therefore,
[tex]f(x) = 7-x^2 \implies y = 7-x^2 \implies x = 7-y^2 \implies y = \pm \sqrt{ 7-x}[/tex]
[tex]\therefore f'(x) = \pm \sqrt{ 7-x}[/tex]
[tex]f'(-1) = \pm \sqrt{ 7-(-1)} = \pm \sqrt{8} = \pm \sqrt{2^3 } = \pm 2\sqrt{2}[/tex]
