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Solutions to exam 1 of math 105a, focusing on derivatives and limits. It includes finding the greatest and least values of functions, using limit definitions to find derivatives, and analyzing the behavior of functions at discontinuities.
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Math 105a Exam 1 Solutions
(a) f is greatest at D (b) f is least at A (c) f 0 is greatest at A (d) f 0 is least at C (e) f 00 is greatest at D (f) f 00 is least A (g) f is increasing everywhere except C since f 0 (x) = 0 at x =C. (h) f is increasing most rapidly at A since f 0 (x) is greatest there.
f 0 (x) = lim h! 0 f^ (x^ +^ h h)^ ^ f^ (x)= lim h! 0 x+^5 h ^ x^5 h = lim h! 0
5 x 5(x + h) xh(x + h)
= lim h! 0 xh(^ x^5 +h h) = lim h! 0 x(x^ +^5 h) = (^) x^52
f 0 (x) = 3x^ ln 3 26 x + (^) x^12
(b) f (x) = 5x^
x^3 + 8x^7
f 0 (x) = 5x^ ln 5
x^3 + 8x^7
3 x^2 + 56x^6
= 5xx^2
8 x^5 ln 5 + 56x^4 + x ln 5 + 3
(c) f (x) = e
x (^) + x 1 x^2 f 0 (x) = (e
x (^) + x (^) ln )(1 x (^2) ) (ex (^) + x) ( 2 x) (1 x^2 )^2 (d) f (x) =
q 1 + p x =
1 + x^1 =^2
f 0 (x) =^12
1 + x^1 =^2
12 x ^1 =^2 = 1 4
p x
p 1 + px
3 x^2 ; if x 0 10(x + 3) (x + 3)(x + 4) ;^ if^ x <^0
Note: (^) (x10( + 3)(x^ + 3)x + 4) = (^) x^10 + 4 provided x 6 = 3
(a) (^) xlim! 0 f (x) = (^) xlim! 0 x 10 + 4 =^104 =^52 (e) (^) x! lim 3 f (x) = (^) x! lim 3 x 10 + 4 = 10
(b) (^) xlim! 0 + f (x) = (^) xlim! 0 + 3 x^2 = 0 (f) (^) x! lim 3 + f (x) = (^) x! lim 3 + x^10 + 4 = 10
(c) (^) x! lim 4 f (x) = (^) x! lim 4 x 10 + 4 = 1 (g) f (0) = 3(0)^2 = 0
(d) (^) x! lim 4 + f (x) = (^) x! lim 4 + x^10 + 4 = 1 (h) f ( 3) is unde ned
Using these limits we can classify all points of discontinuity as either removable or non-removable. First notice that the function f is not continuous at x = 0, x = 3, and x = 4. In particular, (i) f has a removable discontinuity at x = 3 since (^) xlim! 3 f (x) = 10 even though f ( 3) is unde ned. (ii) f has a non-removable discontinuity at x = 4 since (^) x! lim 4 f (x) = 1 and (^) x! lim 4 + f (x) = 1. In other words, the line x = 4 is a vertical asymptote for the graph of f. (ii) f has a non-removable jump discontinuity at x = 0 since (^) xlim! 0 f (x) =^52 and (^) xlim! 0 f (x) = 0.
The graph is shown below.
(a) f 0 (x) = 3 x^2 + 12 = 3(x + 2)(x 2) and f 0 (x) = 0 when x = 2. Creating a sign chart shows that f is decreasing on ( 1; 2) and (2; 1 ).
(b) f 00 (x) = 6 x > 0 if x < 0. Therefore, f is concave up on ( 1; 0).