Question
The equation of the tangent to the curve $y=$ $(2 x-1) e^{2(1-x)}$ at the point of its maximum is(a) $y=1$(b) $x=1$(c) $x+y=1$(d) $x-y=-1$
Step 1
The function is $y=(2x-1)e^{2(1-x)}$. Using the product rule and chain rule, we get \[y' = (2)e^{2(1-x)} - (2x-1)(2)e^{2(1-x)}.\] Show more…
Show all steps
Your feedback will help us improve your experience
Aman Gupta and 100 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Find the equation of the tangent to the curve $y=(2 x-1) e^{2(1-x)}$ at the point of its maximum.
The Tangent and Normal
Level III
$f^{\prime}(x)= \begin{cases}-2 x, & x<0 \\ 2 x, & x \geq 0\end{cases}$ eqn of tangent $\rightarrow \quad y-y_{1}=2 x_{1}\left(x-x_{1}\right), x \geq 0$ $y-y_{2}=-2 x_{2}\left(x-x_{2}\right) x<0$
(a) Find all points on $y^{2}+x y+x^{2}=1$ with $x=1$ (b) Find $d y / d x$ for $y^{2}+x y+x^{2}=1$ (c) Find the slope of the tangent line to $y^{2}+x y+x^{2}=1$ at each point with $x=1$
Short-Cuts to Differentiation
Implicit Functions
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD