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CBSE Class 12 Math 2022 Term I Solved Paper

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A function f:R→Rf: \mathbf{R} \rightarrow \mathbf{R} is defined as f(x)=x3+1f(x)=x^3+1. Then the function has
Solution:     👈: Video Solution
Explanation: Given, f(x)=x3+1f(x)=x^3+1
∴  f′(x)  =3x2   and   f′′(x)=6x\therefore \; f'(x) \; = 3 x^2 \; \text{ and } \; f''(x) = 6 x
   Put   f′(x)  =0\; \text{ Put } \; f'(x) \; = 0
⇒3x2  =0⇒x=0\Rightarrow 3 x^2 \; =0 \Rightarrow x=0
   At   x  =0,f′′(x)=0\; \text{ At } \; x \; =0, f''(x)=0
Thus, f(x)f(x) has neither maximum value nor minimum value.
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