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Math Problems
Precalculus
Find equations of tangent lines using limits
Let
f
f
f
be a function such that
f
(
−
1
)
=
3
f(-1) = 3
f
(
−
1
)
=
3
and
f
′
(
−
1
)
=
5
f'(-1) = 5
f
′
(
−
1
)
=
5
. Let
g
g
g
be the function
g
(
x
)
=
2
x
3
g(x) = 2x^3
g
(
x
)
=
2
x
3
. Let
F
F
F
be a function defined as
F
(
x
)
=
f
(
x
)
g
(
x
)
F(x) = \frac{f(x)}{g(x)}
F
(
x
)
=
g
(
x
)
f
(
x
)
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1
3
x
+
3
y
=
4
2
x
−
4
=
2
y
\begin{aligned} \dfrac{1}{3}x + 3y &= 4 \ 2x - 4 &= 2y \end{aligned}
3
1
x
+
3
y
=
4
2
x
−
4
=
2
y
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Find the equation of the tangent line to
k
(
x
)
=
x
2
k(x) = x^2
k
(
x
)
=
x
2
at
x
=
6
x = 6
x
=
6
.
\newline
Write your answer in point-slope form using integers and fractions. Simplify any fractions.
\newline
y
−
‾
=
‾
(
x
−
‾
)
y - \underline{\quad} = \underline{\quad}(x - \underline{\quad})
y
−
=
(
x
−
)
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Find the equation of the tangent line to
k
(
x
)
=
x
2
k(x) = x^2
k
(
x
)
=
x
2
at
x
=
7
x = 7
x
=
7
.
\newline
Write your answer in point-slope form using integers and fractions. Simplify any fractions.
\newline
y
−
‾
=
‾
(
x
−
‾
)
y - \underline{\quad} = \underline{\quad}(x - \underline{\quad})
y
−
=
(
x
−
)
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4
y
−
3
x
=
40
4y - 3x = 40
4
y
−
3
x
=
40
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Find the equation of the normal to the curve
y
=
x
−
2
2
x
+
1
y=\frac{x-2}{2 x+1}
y
=
2
x
+
1
x
−
2
at the point where the curve cuts the
x
x
x
-axis.
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Given an equation
6
+
4
x
−
2
x
2
=
0
6 + 4x - 2x^2 = 0
6
+
4
x
−
2
x
2
=
0
. (a) Express
6
+
4
x
−
2
x
2
6 + 4x - 2x^2
6
+
4
x
−
2
x
2
in the form
a
−
(
x
+
b
)
2
a - (x + b)^2
a
−
(
x
+
b
)
2
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A curve in the plane is defined parametrically by the equations
\newline
x
=
t
3
+
t
x=t^{3}+t
x
=
t
3
+
t
and
\newline
y
=
t
4
+
2
t
2
y=t^{4}+2t^{2}
y
=
t
4
+
2
t
2
. An equation of the line tangent to the curve at
\newline
t
=
1
t=1
t
=
1
is
\newline
(A)
y
=
2
x
\text{(A)}\ y=2x
(A)
y
=
2
x
\newline
(B)
y
=
8
x
\text{(B)}\ y=8x
(B)
y
=
8
x
\newline
(C)
y
=
2
x
−
1
\text{(C)}\ y=2x-1
(C)
y
=
2
x
−
1
\newline
(D)
y
=
4
x
−
5
\text{(D)}\ y=4x-5
(D)
y
=
4
x
−
5
\newline
(E)
y
=
8
x
+
13
\text{(E)}\ y=8x+13
(E)
y
=
8
x
+
13
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For the function
f
(
x
)
=
x
2
−
10
f(x)=x^{2}-10
f
(
x
)
=
x
2
−
10
, find the slope of the secant line between
x
=
−
6
x=-6
x
=
−
6
and
x
=
1
x=1
x
=
1
.
\newline
Answer:
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For the function
f
(
x
)
=
x
2
−
10
f(x)=x^{2}-10
f
(
x
)
=
x
2
−
10
, find the slope of the secant line between
x
=
−
1
x=-1
x
=
−
1
and
x
=
2
x=2
x
=
2
.
\newline
Answer:
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The exponential function
f
f
f
is graphed in the
x
y
x y
x
y
-plane. As
x
x
x
increases by
1
,
y
1, y
1
,
y
increases by a factor of
3
3
3
. Which of the following could be
f
f
f
?
\newline
Choose
1
1
1
answer:
\newline
(A)
f
(
x
)
=
(
1
3
)
x
f(x)=\left(\frac{1}{3}\right)^{x}
f
(
x
)
=
(
3
1
)
x
\newline
(B)
f
(
x
)
=
(
1
3
)
x
+
3
f(x)=\left(\frac{1}{3}\right)^{x}+3
f
(
x
)
=
(
3
1
)
x
+
3
\newline
(C)
f
(
x
)
=
3
x
+
2
f(x)=3^{x}+2
f
(
x
)
=
3
x
+
2
\newline
(D)
f
(
x
)
=
2
(
3
)
x
f(x)=2(3)^{x}
f
(
x
)
=
2
(
3
)
x
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What is the area of the region between the graphs of
f
(
x
)
=
x
+
1
f(x)=\sqrt{x+1}
f
(
x
)
=
x
+
1
and
g
(
x
)
=
2
x
−
4
g(x)=2 x-4
g
(
x
)
=
2
x
−
4
from
x
=
0
x=0
x
=
0
to
x
=
3
x=3
x
=
3
?
\newline
Choose
1
1
1
answer:
\newline
(A)
23
3
\frac{23}{3}
3
23
\newline
(B)
5
3
\frac{5}{3}
3
5
\newline
(C)
14
3
\frac{14}{3}
3
14
\newline
(D)
−
3
-3
−
3
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Consider the curve given by the equation
3
x
2
+
y
4
+
6
x
=
253
3 x^{2}+y^{4}+6 x=253
3
x
2
+
y
4
+
6
x
=
253
. It can be shown that
d
y
d
x
=
−
6
(
x
+
1
)
4
y
3
.
\frac{d y}{d x}=\frac{-6(x+1)}{4 y^{3}}.
d
x
d
y
=
4
y
3
−
6
(
x
+
1
)
.
\newline
Write the equation of the horizontal line that is tangent to the curve and is above the
x
x
x
-axis.
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What is the area of the region between the graphs of
f
(
x
)
=
x
2
+
12
x
f(x)=x^{2}+12 x
f
(
x
)
=
x
2
+
12
x
and
g
(
x
)
=
3
x
2
+
10
g(x)=3 x^{2}+10
g
(
x
)
=
3
x
2
+
10
from
x
=
1
x=1
x
=
1
to
x
=
4
x=4
x
=
4
?
\newline
Choose
1
1
1
answer:
\newline
(A)
77
77
77
\newline
(B)
64
3
\frac{64}{3}
3
64
\newline
(C)
18
18
18
\newline
(D)
45
45
45
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The graphs of the functions
f
(
x
)
=
sin
(
x
)
f(x)=\sin (x)
f
(
x
)
=
sin
(
x
)
and
g
(
x
)
=
1
2
g(x)=\frac{1}{2}
g
(
x
)
=
2
1
intersect at
2
2
2
points on the interval
0
<
x
<
π
0<x<\pi
0
<
x
<
π
.
\newline
What is the area of the region bound by the graphs of
f
(
x
)
f(x)
f
(
x
)
and
g
(
x
)
g(x)
g
(
x
)
between those points of intersection ?
\newline
Choose
1
1
1
answer:
\newline
(A)
π
3
\frac{\pi}{3}
3
π
\newline
(B)
π
2
\frac{\pi}{2}
2
π
\newline
(C)
2
−
π
2
2-\frac{\pi}{2}
2
−
2
π
\newline
(D)
3
−
π
3
\sqrt{3}-\frac{\pi}{3}
3
−
3
π
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