Your third friend just got her derivatives test back and can’t understand why that teacher of hers took off so many points for the following: lim h → 0 f ( x + h ) − f ( x ) h = lim h → 0 f ( x ) + h − f ( x ) h = lim h → 0 f ( x ) + h − f ( x ) h Now cancel the h = lim h → 0 f ( x ) − f ( x ) Canceled the f ( x ) = 0. X WRONG − 25 What was wrong with your answer? (There may be more than one error.)
Your third friend just got her derivatives test back and can’t understand why that teacher of hers took off so many points for the following: lim h → 0 f ( x + h ) − f ( x ) h = lim h → 0 f ( x ) + h − f ( x ) h = lim h → 0 f ( x ) + h − f ( x ) h Now cancel the h = lim h → 0 f ( x ) − f ( x ) Canceled the f ( x ) = 0. X WRONG − 25 What was wrong with your answer? (There may be more than one error.)
Solution Summary: The author explains the error in the expression cundersethto 0mathrmlim
Your third friend just got her derivatives test back and can’t understand why that teacher of hers took off so many points for the following:
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
=
lim
h
→
0
f
(
x
)
+
h
−
f
(
x
)
h
=
lim
h
→
0
f
(
x
)
+
h
−
f
(
x
)
h
Now cancel the
h
=
lim
h
→
0
f
(
x
)
−
f
(
x
)
Canceled the
f
(
x
)
=
0.
X WRONG
−
25
What was wrong with your answer? (There may be more than one error.)
Use Euler's method to numerically integrate
dy
dx
-2x+12x² - 20x +8.5
from x=0 to x=4 with a step size of 0.5. The initial condition at x=0 is y=1. Recall
that the exact solution is given by y = -0.5x+4x³- 10x² + 8.5x+1
Find an equation of the line tangent to the graph of f(x) = (5x-9)(x+4) at (2,6).
Find the point on the graph of the given function at which the slope of the tangent line is the given slope.
2
f(x)=8x²+4x-7; slope of the tangent line = -3
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.
Differential Equation | MIT 18.01SC Single Variable Calculus, Fall 2010; Author: MIT OpenCourseWare;https://www.youtube.com/watch?v=HaOHUfymsuk;License: Standard YouTube License, CC-BY