1. Find values of a,b,c, and d such that g(x) = a(x^3)+b(x^2)+cx+d has a local maximum at (2,4) and a local minimum at (0,0)
2. Need someone to explain this. Graph a function f(x) for which:
f(3)=f(5)=0
f'(x)<0 if x<4
f'(4)=0
f'(x)>0 if x>4
f"(x)>0 for all x
I know there is concave up/down, zeros and turning points involved but i'm just unsure how to graph it as listed.
3. Find the equation of the tangent to the curve y=e^(2x-1) at the point where x=2.
So to find the derivative of y=e^(2x-1), after I use log/ln then what?
On the second, sketch the points you know, and the shapes, then connect them.
on the first..
Take the first and second derivatives. Set the first to zero, an solve for roots at x=2, 0. You can use the second derivative to assist.
On the third. You know the slope (the value of the derivative at x=2)
so y=mx + b you know y, m, x at the point of tangency, solve for b.
1. To find the values of a, b, c, and d such that the function g(x) = a(x^3) + b(x^2) + cx + d has a local maximum at (2,4) and a local minimum at (0,0), we can use the properties of derivative.
First, let's find the first derivative of g(x) with respect to x:
g'(x) = 3ax^2 + 2bx + c
Since there is a local maximum at (2,4), the derivative at x=2 should be zero:
g'(2) = 3a(2^2) + 2b(2) + c = 0 ---> 12a + 4b + c = 0 ----(1)
Similarly, since there is a local minimum at (0,0), the derivative at x=0 should also be zero:
g'(0) = 3a(0^2) + 2b(0) + c = 0 ---> c = 0 ----(2)
Now, let's find the second derivative of g(x):
g''(x) = 6ax + 2b
To have a local maximum at x=2, the second derivative at x=2 should be negative:
g''(2) = 6a(2) + 2b < 0 ---> 12a + 2b < 0 ----(3)
To have a local minimum at x=0, the second derivative at x=0 should be positive:
g''(0) = 6a(0) + 2b > 0 ---> 2b > 0 ----(4)
From equation (4), we can see that b must be greater than zero.
Now, we have equations (1)-(4) to solve for the values of a, b, c, and d. With the given conditions, solving these equations simultaneously will give us the desired values for the coefficients of g(x).
2. To graph a function f(x) that satisfies the given conditions:
- f(3) = f(5) = 0 indicates that the graph crosses the x-axis at x=3 and x=5. So, the graph will have x-intercepts at x=3 and x=5.
- f'(x) < 0 if x < 4 indicates that the graph is decreasing on the interval (-∞, 4).
- f'(4) = 0 indicates that the graph has a critical point at x=4, where the slope changes from negative to positive.
- f'(x) > 0 if x > 4 indicates that the graph is increasing on the interval (4, ∞).
- f''(x) > 0 for all x indicates that the graph is concave up everywhere.
Based on these conditions, you can start by plotting the x-intercepts at x=3 and x=5. Then, draw a decreasing curve on the interval (-∞, 4) and an increasing curve on the interval (4, ∞). Make sure the slope is 0 at x=4. Finally, draw the graph in a concave up shape to satisfy the condition of f''(x) > 0 for all x.
3. To find the equation of the tangent to the curve y = e^(2x-1) at the point where x=2, you need to find the derivative of y with respect to x, evaluate it at x=2 to find the slope of the tangent, and then use the slope-intercept form of a line to find the equation of the tangent line.
First, find the derivative of y with respect to x:
dy/dx = d/dx(e^(2x-1)) = e^(2x-1) * d/dx(2x-1) = 2e^(2x-1)
Evaluate the derivative at x=2 to find the slope of the tangent:
dy/dx(x=2) = 2e^(2*2-1) = 2e^3
Now, you have the slope (m = 2e^3) and a point on the line (x=2, y=e^(2*2-1)), which can be expressed as (2, e^3). Use the point-slope form to find the equation of the tangent line:
y - y1 = m(x - x1)
y - e^3 = 2e^3(x - 2)
Simplify the equation to get the equation of the tangent line to the curve y = e^(2x-1) at the point where x=2.