Video: APCALC03AB-P1A-Q01-906132783279

If 𝑓(π‘₯) = |π‘₯⁡|, what is the value of lim_(π‘₯ β†’ βˆ’1) 𝑓′(π‘₯).

02:43

Video Transcript

If 𝑓 of π‘₯ equals the absolute value of π‘₯ to the fifth power, what is the value of the limit as π‘₯ tends to negative one of 𝑓 prime of π‘₯.

Let’s have a look at this function we’ve been given. This is an absolute value function, as indicated by these two vertical lines, which returns the value of a function without its sign. So, for example, both five and negative five get the value five. We can choose instead to express 𝑓 of π‘₯ without this notation if we define it as a piecewise function. For values of π‘₯ greater than or equal to zero, raising them to the fifth power will give another number greater than or equal to zero, i.e., a nonnegative value. And therefore, the absolute value of π‘₯ to the fifth power is just the value of π‘₯ to the fifth power itself.

However, for values of π‘₯ less than zero, raising them to the fifth power will give a negative result because we’re raising them to an odd power. And therefore, the absolute value of π‘₯ to the fifth power will be the negative of π‘₯ to the fifth power. So we can define our function 𝑓 of π‘₯ as a piecewise function. It’s equal to negative π‘₯ to the fifth power when π‘₯ is less than zero or π‘₯ to the fifth power when π‘₯ is greater than or equal to zero. To find the limit as π‘₯ tends to negative one of 𝑓 prime of π‘₯, that’s the first derivative of this function, we just need to consider the piece of this function that includes π‘₯ equals negative one. And as negative one is less than zero, it will be the top piece of our function.

We can differentiate this piece of our function using the power rule of differentiation. And we see that the limit as π‘₯ tends to negative one of 𝑓 prime of π‘₯ will be equal to the limit as π‘₯ tends to negative one of negative five π‘₯ to the fourth power. We have brought the power down and then reduced the power by one.

We can now evaluate this limit as π‘₯ tends to negative one by substituting negative one for π‘₯. And it gives negative five multiplied by negative one to the fourth power. Negative one to the fourth power is positive one because this is an even power, so we get a positive result. We have negative five multiplied by one which is equal to negative five. So by rewriting our absolute value function as a piecewise function, we were able to differentiate the relevant part of it and hence find the limit as π‘₯ tends to negative one of its first derivative, which is equal to negative five.

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