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Question: How do you show the function \( f\left( x \right)={{\left( x+2{{x}^{3}} \right)}^{4}} \) is continuo...

How do you show the function f(x)=(x+2x3)4f\left( x \right)={{\left( x+2{{x}^{3}} \right)}^{4}} is continuous at the given number a=−1a=-1 ?

Explanation

Solution

Hint : We have to find the continuity of the given function at a certain point a=−1a=-1 . We use the points and use their close values to find the different functions that will be available to operate. We take the final conclusion depending on the equality of the theorems lim⁡x→a+ f(x)=lim⁡x→a− f(x)=f(a)\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right) for continuity.

Complete step-by-step answer :
We have to show the continuity and the differentiability of the given function at certain points.
For the function if the condition lim⁡x→a+ f(x)=lim⁡x→a− f(x)=f(a)\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right) satisfies then it will continuous.
At a=−1a=-1 , we break the point in three parts where a=−1,−1+,−1−a=-1,-{{1}^{+}},-{{1}^{-}} .
We check the values of the function at those points.
At a=−1−a=-{{1}^{-}} , we have
lim⁡x→−1− f(x)=f(−1)=[(x+2x3)4]x=−1=(−3)4=81\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=f\left( -1 \right)={{\left[ {{\left( x+2{{x}^{3}} \right)}^{4}} \right]}_{x=-1}}={{\left( -3 \right)}^{4}}=81
At a=−1+a=-{{1}^{+}} , we have
lim⁡x→−1+ f(x)=f(−1)=[(x+2x3)4]x=−1=(−3)4=81\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( -1 \right)={{\left[ {{\left( x+2{{x}^{3}} \right)}^{4}} \right]}_{x=-1}}={{\left( -3 \right)}^{4}}=81
We also have f(−1)=[(x+2x3)4]x=−1=(−3)4=81f\left( -1 \right)={{\left[ {{\left( x+2{{x}^{3}} \right)}^{4}} \right]}_{x=-1}}={{\left( -3 \right)}^{4}}=81.
Therefore, f(x)f\left( x \right) is continuous at a=−1a=-1 as lim⁡x→−1− f(x)=f(−1)=lim⁡x→−1+ f(x)=81\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=f\left( -1 \right)=\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=81.
The continuity of the functions is also essential for the differentiability of the function.
So, the correct answer is “81”.

Note : This type of differentiability checking is called differentiability of piecewise function. A piecewise function is differentiable at a point if both of the pieces have derivatives at that point, and the derivatives are equal at that point.