derivative of 4x^3+12x^2-14 is
Answers
Step-by-step explanation:
We can find the slope of a line tangent to a curve at a point by evaluating the derivative of the function at that point.
We are given the function
f
(
x
)
=
4
x
3
+
12
x
2
+
9
x
+
7
=
f
(
x
)
=
4
x
3
+
12
x
2
+
9
x
1
+
7
x
0
Using the power rule, let's now compute the derivative of
f
(
x
)
f
'
(
x
)
=
(
3
⋅
4
x
2
)
+
(
2
⋅
12
x
)
+
(
1
⋅
9
x
0
)
+
0
⋅
7
x
−
1
f
'
(
x
)
=
12
x
2
+
24
x
+
9
We can now find the slope of
f
(
x
)
at
x
=
−
3
by substituting this value into
f
'
(
x
)
f
'
(
−
3
)
=
12
(
−
3
)
2
+
24
(
−
3
)
+
9
f
'
(
−
3
)
=
12
(
9
)
+
24
(
−
3
)
+
9
f
'
(
−
3
)
=
108
−
72
+
9
f
'
(
−
3
)
=
45
(slope of the tangent line at
x
=
−
3
Now that we have a slope for the tangent line, we need to identify a point on the line.
We know the tangent line touches the function
f
(
x
)
at the point
x
=
−
3
, so let's find the value of
f
(
x
)
at this point:
f
(
−
3
)
=
4
(
−
3
)
3
+
12
(
−
3
)
2
+
9
(
−
3
)
+
7
f
(
−
3
)
=
4
(
−
27
)
+
12
(
9
)
+
9
(
−
3
)
+
7
f
(
−
3
)
=
−
108
+
108
−
27
+
7
f
(
−
3
)
=
−
20
So we know the tangent line goes through the point
(
−
3
,
−
20
)
Finally, we can use the point-slope formula for a line to find the equation of the tangent line.
y
=
m
x
+
b
To find the value of
b
, substitute the values we have calculated for the point and slope of the tangent line:
(
−
20
)
=
(
45
)
(
−
3
)
+
b
−
20
=
−
135
+
b
b
=
115
So our final answer for the equation of the tangent line is:
y=45x+115
please mark it as brainliest because I tried my level best