1. On what interval is the function f(x)=x^3-4x^2+5x concave upward?
2. For what values of x dies the graph of f(x)=2x^3-3x^2-6x+87 have a horizontal tangent?
Is there no solution because the equation can't be factored?
3. At what point on the curve y=1+2e^x - 3x is the tangent line parallel to the line 3x-y=5?
#1
f' = 3x^2-8x+5
f" = 6x-8
f is concave up when f" > 0
#2
f' = 6x^2-6x-6
maybe it cannot be factored, but since the discriminant is positive, it has real solutions. Use the quadratic formula to find them.
#3 you want the slope to be 3.
y' = 2e^x - 3
so,
2e^x - 3 = 3
2e^x = 6
e^x = 3
x = ln 3
y(ln3) = 1 + 6 - 3ln3 = 7-ln27
So, the tangent line is
y = 3(x-ln3) + 7-ln27
See the graph at
http://www.wolframalpha.com/input/?i=plot+y%3D1%2B2e^x+-+3x%2C+y%3D3%28x-ln3%29+%2B+7-ln27
1. Ah, the timeless question of concavity! Well, to determine where the function f(x)=x^3-4x^2+5x is concave upward, we'll have to find where the second derivative is positive. Mathematically speaking, that's where the fun begins. So, let's take the second derivative. Ready for some math magic? Here it goes: f''(x) = 6x - 8. Now, solving the inequality 6x - 8 > 0, we find x > 8/6 or, in simpler terms, x > 4/3. So, the function is concave upward on the interval x > 4/3. Now, let's hope the math police don't give me a ticket for excessive humor!
2. Ah, horizontal tangents – the sneaky ones, aren't they? To find where the graph of f(x)=2x^3-3x^2-6x+87 has horizontal tangents, we need to find where the derivative equals zero. So, let's differentiate! After applying some mathematical gymnastics, we find f'(x) = 6x^2 - 6x - 6. To find x values where the derivative equals zero, we set 6x^2 - 6x - 6 = 0 and use the quadratic formula (or try a fancy dance routine). After some shaking and spinning, we find x = -0.5 and x = 2. Phew, finally! So, the graph has horizontal tangents at x = -0.5 and x = 2. It's time to break out the confetti and celebrate solving this tangential mystery!
3. Ah, the Parallel Line Police are on the case! So, to find where the tangent line to the curve y=1+2e^x - 3x is parallel to the line 3x-y=5, we need to find where the slopes are equal. Ready for some math rollercoaster? Okay, the slope of the tangent line to the curve y=1+2e^x - 3x is equal to the derivative of that function, which is 2e^x - 3. The slope of the line 3x-y=5 can be found by rearranging it to y = 3x - 5, giving us a slope of 3. To find where these slopes are equal, we need to solve the equation 2e^x - 3 = 3. Now, we just need to do some mathematical gymnastics (or hire a juggling magician). After some calculations and clever moves, we find x = ln(3/2). So, the tangent line is parallel to the line 3x-y=5 at the point (ln(3/2), 1+2e^(ln(3/2)) - 3ln(3/2) ). Those lines are in parallel party mode now!
1. To determine the intervals on which the function f(x) = x^3 - 4x^2 + 5x is concave upward, we need to find the second derivative and analyze its sign.
First, find the first derivative by differentiating f(x):
f'(x) = 3x^2 - 8x + 5
Then, find the second derivative by differentiating f'(x):
f''(x) = 6x - 8
To determine the intervals on which f(x) is concave upward, we need to solve the inequality f''(x) > 0.
Setting f''(x) > 0:
6x - 8 > 0
Solving for x:
6x > 8
x > 8/6
x > 4/3
Therefore, the function f(x) = x^3 - 4x^2 + 5x is concave upward on the interval (4/3, ∞).
2. To find the values of x where the graph of f(x) = 2x^3 - 3x^2 - 6x + 87 has a horizontal tangent, we need to set the first derivative equal to zero and solve for x.
First, find the first derivative by differentiating f(x):
f'(x) = 6x^2 - 6x - 6
Set f'(x) equal to zero:
6x^2 - 6x - 6 = 0
This quadratic equation can be factored as follows:
(2x + 3)(3x - 2) = 0
Setting each factor equal to zero and solving for x:
2x + 3 = 0 --> x = -3/2
3x - 2 = 0 --> x = 2/3
Therefore, the graph of f(x) = 2x^3 - 3x^2 - 6x + 87 has horizontal tangents at x = -3/2 and x = 2/3.
3. To find the point on the curve y = 1 + 2e^x - 3x where the tangent line is parallel to the line 3x - y = 5, we need to find the slope of the given line and set it equal to the derivative of the function at that point.
First, rearrange the equation 3x - y = 5 to the slope-intercept form:
y = 3x - 5
The slope of this line is 3.
Next, find the derivative of the function y = 1 + 2e^x - 3x:
dy/dx = 2e^x - 3
Set the derivative equal to the slope of the line:
2e^x - 3 = 3
Solve for e^x:
2e^x = 6
e^x = 3
Take the natural logarithm of both sides to solve for x:
x = ln(3)
Substitute this value of x back into the original function to find the y-coordinate:
y = 1 + 2e^x - 3x
y = 1 + 2e^(ln(3)) - 3(ln(3))
y = 1 + 2(3) - 3ln(3)
y = 7 - 3ln(3)
Therefore, the point on the curve y = 1 + 2e^x - 3x where the tangent line is parallel to 3x - y = 5 is (ln(3), 7 - 3ln(3)).
1. To determine the interval where the function f(x) = x^3 - 4x^2 + 5x is concave upward, you need to find the second derivative of the function and analyze its sign.
First, find the first derivative by differentiating the function:
f'(x) = 3x^2 - 8x + 5
Now, find the second derivative by differentiating f'(x) with respect to x:
f''(x) = 6x - 8
To determine the interval where the function is concave upward, set the second derivative greater than zero and solve for x:
6x - 8 > 0
Solving the inequality, you get:
x > 8/6
x > 4/3
Therefore, the function f(x) = x^3 - 4x^2 + 5x is concave upward for x greater than 4/3.
2. To find the values of x where the graph of f(x) = 2x^3 - 3x^2 - 6x + 87 has a horizontal tangent, you need to find the derivative of the function and set it equal to zero.
First, find the derivative of f(x):
f'(x) = 6x^2 - 6x - 6
Now, set the derivative equal to zero and solve for x:
6x^2 - 6x - 6 = 0
You can attempt to factor this quadratic equation or use the quadratic formula to find the solutions. If the equation cannot be factored easily, you can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 6, b = -6, and c = -6. Plugging in these values, you can solve for x.
Therefore, the equation can indeed be factored or solved using the quadratic formula.
3. To find the point on the curve y = 1 + 2e^x - 3x where the tangent line is parallel to the line 3x - y = 5, you need to find the derivative of the given curve and set it equal to the slope of the given line.
First, find the derivative of the curve:
dy/dx = 2e^x - 3
Next, set the derivative equal to the slope of the given line, which is the coefficient of x in the equation 3x - y = 5:
2e^x - 3 = 3
Solve the equation for x by isolating the exponential term:
2e^x = 6
e^x = 3
Take the natural logarithm of both sides to solve for x:
x = ln(3)
Substitute the value of x back into the original equation to find the corresponding y-coordinate:
y = 1 + 2e^(ln(3)) - 3(ln(3))
y = 1 + 2(3) - 3(ln(3))
y = 1 + 6 - 3(ln(3))
So, the point where the tangent line is parallel to the line 3x - y = 5 is (ln(3), 7 - 3(ln(3))).