Let f be a twice-differentiable function defined on the interval -1.2 less than or equal to x less than or equal to 3.2 with f(1)=2. The graph of f', the derivative of f, is shown on the right. The graph of f' crosses the x-axis at x=-1 and x=3 and has a horizontal tangent at x=2. Let g be the function given by g(x)=e^(f(x)).
View graph here: h t t p : / / e t s . o p e n s t u d y . c o m / u p d a t e s / a t t a c h m e n t s / 5 0 d 1 e f f a e 4 b 0 6 9 a b b b 7 1 0 d 3 9 - b l a d e r u n n e r 1 1 2 2 - 1 3 5 5 9 3 5 7 7 5 8 1 7 - g r a p h . p n g
1. Write an equation for the line tangent to the graph of g at x=1. 2. For -1.2 is less than or equal to x is less than or equal to 3.2, find all values of x at which g has a local maximum. Justify your answer. 3. The second derivative of g is g''(x)=x^(f(x)) [(f'(x))^2 + f''(x)]. Is g''(-1) positive, negative, or zero? Justify your answer. 4. Find the average rate of change of g', the derivative of g, over the interval [1,3].
1.
g(x) = e^f(x)
g'(x) = e^f(x) f'(x)
g'(1) = e^f(1) f'(1) = e^2 (-4) = -4e^2
g(1) = e^f(1) = e^2
so, you want the line through (1,e^2) with slope -4e^2:
y-e^2 = -4e^2 (x-1)
2.
g has a max where g' = 0
since e^f > 0 for all x, g'=0 when f'=0. So, g has a max/min at x = -1 or x=3
Since f''<0 at x=-1, g is a max at x = -1.
3.
e^f > 0
f'(-1) = 0
f''(-1) < 0
so, g''(-1) < 0
4.
g'(3) = e^f(3) f'(3) = 0
g'(1) = -4e^2 as above
avg change is
(g'(3)-g'(1))/(3-1) = (0- -4e^2)/2 = 2e^2
Two particles move along the x -axis. For 0 is less than or equal to t is less than or equal to 6, the position of particle P at time t is given by p(t)=2cos((pi/4)t), while the position of particle R at time t is given by r(t)=t^3 -6t^2 +9t+3.
1. For 0 is less than or equal to t is less than or equal to 6, find all times t during which particle R is moving to the left. 2. for 0 is less than or equal to t is less than or equal to 6, find all times t during which the two particles travel in opposite directions. 3. Find the acceleration of particles P at time t=3. Is particle P speeding up, slowing down, or doing neither at time t=3? Explain your reasoning. 4. Show that during the interval (1,3), there must be at least one instant when the particle R must have a velocity of -2.
To answer the questions, we need to analyze the given information step by step. Here's how we can approach each question:
1. To find the equation of the tangent line to the graph of g at x = 1, we need to find the slope of the tangent line. The slope of the tangent line is equal to the value of g'(1), the derivative of g at x = 1. We can refer to the graph of f' to determine this value. Since f' has a horizontal tangent at x = 2, we can conclude that f'(1) = 0, which means g'(1) = e^f(1) * f'(1) = 0. Therefore, the slope of the tangent line is 0.
Next, we find the y-coordinate of the point of tangency, which is g(1). Using the equation g(x) = e^f(x), we can substitute x = 1 to find g(1) = e^f(1) = e^2.
Now, we have the slope (0) and a point on the line (1, e^2). Using the point-slope form of the equation of a line, we can write the equation of the tangent line as y - e^2 = 0 * (x - 1), which simplifies to y = e^2.
2. To find all values of x at which g has a local maximum, we need to locate the critical points of g. Critical points occur where g'(x) = 0 or where g'(x) is undefined. In this case, g'(x) is the derivative of g, which can be found using the chain rule.
Using the given function g(x) = e^f(x), we differentiate it with respect to x:
g'(x) = e^f(x) * f'(x).
From the graph of f', we know that f'(x) crosses the x-axis at x = -1 and x = 3. Thus, at these points, f'(x) = 0. Since g'(x) = e^f(x) * f'(x), g'(x) will also be 0 at x = -1 and x = 3.
Now, we need to consider the interval -1.2 ≤ x ≤ 3.2 to find any additional critical points. We can see from the graph of f' that there are no other points where f'(x) is undefined or equals zero. Therefore, the critical points are x = -1 and x = 3.
To determine if these critical points are local maxima or minima, we can examine the behavior of g'(x) in the vicinity of these points. However, without more information about the behavior of f(x) on the specified interval, we cannot definitively determine if these critical points are local maxima. Additional information is needed to justify the answer.
3. To determine the sign of g''(-1), we need to evaluate the second derivative of g at x = -1, which is given as g''(x) = x^(f(x)) * [(f'(x))^2 + f''(x)].
Substituting x = -1 into this equation, we get g''(-1) = (-1)^(f(-1)) * [(f'(-1))^2 + f''(-1)].
From the graph of f', we know that f'(x) = 0 at x = -1, and the graph of f'' is not provided. We cannot determine the value of f''(-1) or f(-1) solely from the given information. Thus, without more information about the behavior of f(x) and f''(x), we cannot determine the sign of g''(-1) or justify our answer.
4. To find the average rate of change of g'(x) over the interval [1, 3], we need to evaluate the difference between g'(3) and g'(1) over the interval length, which in this case is 2 (3 - 1).
From the graph of f', we can observe that g'(3) = e^f(3) * f'(3) and g'(1) = e^f(1) * f'(1). However, we do not have specific information about the values of f(3) and f'(3). Without this information, we cannot calculate the average rate of change of g'(x) over the interval [1, 3]. Additional information is required to compute the answer.
In summary, to completely answer these questions, we need additional information about the functions f(x), f''(x), and the behavior of f(x) on the specified interval [-1.2, 3.2]. Without this additional information, we can only make partial conclusions or say that we are unable to determine the answer.