Consider the following function.
f(x) = 4x3 + x2 - 0.05x + 10
Find the two points on the graph at which the tangent line is horizontal. G
To find the points on the graph at which the tangent line is horizontal, we need to find the values of x that make the derivative of the function equal to zero. When the derivative is zero, it means that the slope of the tangent line is zero, which corresponds to a horizontal tangent.
First, let's find the derivative of the function f(x):
f(x) = 4x^3 + x^2 - 0.05x + 10
To find the derivative, we'll differentiate each term with respect to x:
f'(x) = d/dx (4x^3) + d/dx (x^2) - d/dx (0.05x) + d/dx (10)
Using the power rule of differentiation, we get:
f'(x) = 12x^2 + 2x - 0.05
Now, we set the derivative equal to zero to find the critical points:
12x^2 + 2x - 0.05 = 0
This is a quadratic equation, so we can solve it using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 12, b = 2, and c = -0.05. Plugging these values into the quadratic formula, we get:
x = (-2 ± √(2^2 - 4 * 12 * (-0.05))) / (2 * 12)
x = (-2 ± √(4 + 2.4)) / 24
x = (-2 ± √6.4) / 24
Simplifying further, we get:
x = (-2 ± √6.4) / 24
x = (-1 ± √1.6) / 12
Now we have two possible values for x: (-1 + √1.6) / 12 and (-1 - √1.6) / 12.
To find the corresponding y-values, we substitute these x-values back into the original function f(x):
y = f((-1 + √1.6) / 12) and y = f((-1 - √1.6) / 12)
After plugging in these x-values and evaluating the function, we will get the corresponding y-values, giving us the two points on the graph where the tangent lines are horizontal.
To find the points on the graph where the tangent line is horizontal, we need to find the x-values where the derivative of the function is equal to zero.
Step 1: Find the derivative of the function.
The derivative of f(x) = 4x^3 + x^2 - 0.05x + 10 can be found using the power rule for differentiation. Taking the derivative term by term, we get:
f'(x) = 12x^2 + 2x - 0.05
Step 2: Set the derivative equal to zero and solve for x.
To find the x-values where the tangent line is horizontal, we need to find the x-values where the derivative f'(x) is equal to zero. Hence, we set f'(x) = 0 and solve for x:
12x^2 + 2x - 0.05 = 0
Step 3: Solve the quadratic equation.
To solve the quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring is a bit trickier, so we will use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 12, b = 2, and c = -0.05. Plugging these values into the formula, we get:
x = (-(2) ± √((2)^2 - 4(12)(-0.05))) / (2(12))
Simplifying further:
x = (-2 ± √(4 + 2.4)) / 24
x = (-2 ± √6.4) / 24
Step 4: Simplify the solutions.
We can simplify the solutions further by dividing the numerator and the denominator by 2:
x = (-1 ± √1.6) / 12
So, the two x-values where the tangent line is horizontal are:
x₁ = (-1 + √1.6) / 12
x₂ = (-1 - √1.6) / 12
These are the two points on the graph at which the tangent line is horizontal.