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Christian Parkinson GREPrep: Calculus I Practice Problem Solutions 1
Week 1: Calculus I
Practice Problem Solutions
Problem 1. What is the tangent line to the graph of y = x+ex at x = 0?
Solution. The tangent line is given by ℓ(x) = y(0) + y′(0)(x − 0) = 1 + 2x.
(1 +x)α −1
Problem 2. Evaluate lim for α ∈ R
x→0 x
Solution. Using l’Hˆopital’s rule, we see
lim (1 + x)α − 1 = lim α(1+x)α−1 = α.
x→0 x x→0 1
This gives a first order approximation (1 + x)α ∼ 1 + αx when x ≈ 0.
Problem 3. Evaluate lim cos(βx)−1 for β ∈ R.
x→0 2
x
Solution. Using l’Hˆopital’s rule, we see
lim cos(βx)−1 = lim −βsin(βx)
x→0 x2 x→0 2x
−β2cos(βx) β2
=lim =− .
x→0 2 2
Problem 4. Let c > 0. Find the minimum value of f(x) = ex −cx among x ∈ R.
Solution. Setting the derivative to zero shows that extreme points occur when
ex −c = 0 ⇐⇒ x=log(c).
The second derivative of f is always positive so any extreme point is a minimum. Thus the
minimum value is f(log(c)) = c−clog(c).
Problem 5. Let f(x) = |x|+3x2 for x ∈ R. What is f′(−1)?
Solution. In a neighborhood of −1, we have |x| = −x and so f(x) = −x + 3x2. Then
f′(−1) = −1+6(−1) = −7.
−2 −2
Problem 6. Compute the limit lim(x −sin(x) ).
x→0
Christian Parkinson GREPrep: Calculus I Practice Problem Solutions 2
Solution. We see
2 2 2 2 2
−2 −2 sin (x) − x x sin (x) − x
lim(x −sin(x) ) = lim 2 =lim 2
x→0 x→0 x2sin (x) x→0 sin (x) x4
| {z }
→1
sin(x)+x sin(x) − x
=lim
x→0 x x3
| {z }
→2
=2 lim cos(x)−1 = 2lim −sin(x) = −1.
x→∞ 3x2 x→0 6x 3
2
Problem7. Supposethatf(x) = 3x +bx+chasanon-simplerootatx = 2. Whatisf(5)?
Solution. If a quadratic polynomial has a non-simple root at x = 2 then it is a multiple of
(x−2)2. Here
2 2
f(x) = 3(x−2) =⇒ f(5)=3(3) =27.
Problem 8. If f : R → R is continuously differentiable on (−1,4) with f(3) = 5 and
f′(x) ≥ 1 for all x ∈ (−1,4), what is the greatest possible value of f(0)?
Solution. Using the fundamental theorem of calculus, we have
f(0) = f(3)−Z 3f′(x)dx ≤ f(3)−Z 31 dx = 2.
0 0
This bound can be realized if f′(x) ≡ 1 [so f(x) = x + 2].
Problem 9. Compute lim sin(2x) .
x→0 (1 +x)ln(1+x)
Solution. The term 1 is irrelevant since it tends to 1 in the limit. Thus
1+x
lim sin(2x) =lim sin(2x) = lim2(1+x)cos(2x) = 2.
x→0 (1 +x)ln(1+x) x→0 ln(1 + x) x→0
g(x) ′ ′
Problem 10. Let f(x) = e h(x) where h(x) = −g (x)h(x) for all x ∈ R. Which of the
following is necessarily true? (a) f is constant (b) f is linear and non-constant
(c) g is constant (d) g is linear and non-constant (e) none of the above
Solution. Note that
f′(x) = eg(x)g′(x)h(x) + eg(x)h′(x) = eg(x)g′(x)h(x) − eg(x)g′(x)h(x) = 0
Christian Parkinson GREPrep: Calculus I Practice Problem Solutions 3
so f is constant.
Problem 11. Let f(x) = x2+sin(x) for x > 0. Find f′(x).
Solution. The temptation here is to use the power rule or the exponential rule but in the
current form, neither apply since both the base and the exponent depend on x. To fix this,
we write f(x) = e(2+sin(x))log(x). Thus
′ (2+sin(x))log(x) 2 + sin(x) 2+sin(x) 2 + sin(x)
f (x) = e x +cos(x)log(x) =x x +cos(x)log(x) .
R1√ 4 R1√ 4 R1√ 8
Problem 12. Let J = 0 1−x dx, K = 0 1+x dx, L= 0 1−x dx. Order the
numbers J,K,L,1 in increasing order.
Solution. For x ∈ (0,1), we have
4 8 4
1−x <1−x <1<1+x.
Taking the square root and integrating shows that J < L < 1 < K.
Problem 13. Find c ∈ R such that g : R → R satisfies 3x5 +96 = Rxg(t)dt.
c
4
Solution. Differentiating shows that g(x) = 15x . Thus
5 5 5
3x +96=3x −3c =⇒ c=−2.
Problem 14. Define f(0) = 0 and f(x) = |x| for x 6= 0. Compute Z 1 f(x)dx.
x
−1
Solution. The function is odd so the integral on a symmetric range is zero.
Problem 15. Let f(x) = Rx dt2. Find an equation for the tangent line at (2,f(2)).
1 1+t
Solution. The tangent line is given by ℓ(x) = f(2) + f′(2)(x − 2). Here
f(2) = arctan(2)− π
4
and f′(2) = 1 =1 so ℓ(x) = arctan(2)− π + 1(x−2).
2
1+2 5 4 5
Rx 2 2 −1 ′
Problem 16. Let f(x) = 0 cos (t )dt. Find (f ) (y) for y = f(3).
Christian Parkinson GREPrep: Calculus I Practice Problem Solutions 4
Solution. Recall that (f−1)′(y) = 1 . Thus
f′(f−1(y))
−1 ′ 1 2
(f ) (f(3)) = f′(3) = sec (9).
Problem 17. For continuous functions f,g : R → R, define the relation ∼ by f ∼ g iff
lim f(x) = 1.
x→∞ g(x)
Suppose that f ∼ g. Which of these does NOT necessarily follow:
2 2 √ √ f g
(a) f ∼ g (b) f ∼ g (c) e ∼ e (d) f +g ∼ 2g (e) g ∼ f
Solution. The one which does NOT follow is (c) ef ∼ eg. Indeed, put f(x) = x and
g(x) = x−1. Then f ∼ g, but
ef(x)
lim g(x) = lim e = e 6= 1
x→∞e x→∞
so ef 6∼ eg.
Problem 18. Let g(x) = e2x+1. Compute lim g(g(x))−g(e).
x→0 x
Solution. The key is to recognize the limit as (g ◦ g)′(0). Now
(g ◦ g)′(x) = d e2e2x+1+1 = e2e2x+1+1(4e2x+1)
dx
′ 2e+1 2e+2
so (g ◦ g) (0) = e · (4e) = 4e .
Problem19. Supposethatf isdifferentiableatx = x0. Whatis lim f(x0 +h)−f(x0 −h)?
h→0 h
Solution. By adding and subtracting f(x0) in the middle of the numerator, we see that
this limit is 2f′(x0).
Problem 20. Compute the derivative d Z x2 e−t2dt.
dx 0
Solution. By the fundamental theorem of calculus and the chain rule
Z 2
d x
2 4
e−t dt = 2xe−x .
dx 0
3
x
Problem 21. Find the first derivative of f(x) = √ when x > 0.
2 3 4
(6x +1) (x+3)
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