Calculus Volume 1

240

Chapter 3 | Derivatives

Figure 3.17 For the car to move smoothly along the track, the function must be both continuous and differentiable.

Solution For the function to be continuous at x =−10, lim x →10 −

f ( x ) = f (−10). Thus, since

lim x →−10 − 2 −10 b + c =10−10 b + c and f (−10) =5, we must have 10−10 b + c =5. Equivalently, we have c =10 b −5. For the function to be differentiable at −10, f ′(10) = lim x →−10 f ( x )− f (−10) x +10 must exist. Since f ( x ) is defined using different rules on the right and the left, we must evaluate this limit from the right and the left and then set them equal to each other: f ( x ) = 1 10 (−10)

x 2 + bx + c −5 x +10

x +10 = lim x →−10 − 1 10

f ( x )− f (−10)

lim x →−10 −

1 10 x 2 + bx +(10 b −5)−5 x +10 x 2 −100+10 bx +100 b 10( x +10) ( x +10)( x −10+10 b ) 10( x +10)

= lim

Substitute

c =10 b −5.

x →−10 −

= lim

x →−10 −

= lim = b −2.

Factor by grouping.

x →−10 −

We also have

x + 5 x +10

− 1 4

2 −5

f ( x )− f (−10)

lim x →−10 +

x +10 = lim x →−10 +

−( x +10) 4( x +10)

= lim

x →−10 +

= − 1 4 .

and c =10 ⎛

⎞ ⎠ −5= 25 2 .

⎝ 7 4

This gives us b −2= − 1 4 .

Thus b = 7 4

Find values of a and b that make f ( x ) = ⎧ ⎩

⎨ ax + b if x <3 x 2 if x ≥3

3.8

both continuous and differentiable at 3.

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