Let f(x) = 2x^2 + 1
a. Find the derivative f' of f.
b. Find an equation ofthe tangent line to the curve at the point (1,3).
c. Sketch the graph of f.
*Can someone please help me? I have been trying to do my homework for some time now and I am stuck on this problem. I do not even understand what question 1a. is asking. What does f' of f refer to???
i know how to do part B....
I do not know how to do part C -
any help would be great!!! :)
I find it hard to understand that somebody who is studying Calculus would ask this question, and I don't know where to begin to give you an answer.
This is as basic as it can get in terms of first year Calculus.
Did you walk into the wrong math class, and I am not being facicious here.
No worries, I'm here to help you! Let's break down the problem step by step and explain each part to make it easier for you.
a. The notation f' represents the derivative of the function f(x). The derivative gives you the rate at which the function is changing at a specific point. In order to find the derivative f'(x), you need to apply the power rule for derivatives. For a term with the form ax^n (where a is a constant and n is a positive integer), the derivative is given by multiplying the coefficient a by the exponent n and then decreasing the exponent by 1.
In this case, f(x) = 2x^2 + 1. To find f'(x), we need to differentiate each term separately.
The derivative of 2x^2 is obtained as follows:
- Multiply the coefficient 2 by the exponent 2 to get 4.
- Decrease the exponent by 1, resulting in x^1, which is simply x.
Hence, the derivative of 2x^2 is 4x.
The derivative of the constant term 1 is simply 0, as a constant does not change.
Therefore, the derivative of f(x) = 2x^2 + 1 is f'(x) = 4x.
b. To find the equation of the tangent line at the point (1,3), we need to utilize the point-slope form of a linear equation. The tangent line passes through the point (1,3), so we can use the coordinates (x₁, y₁) = (1,3) and the derivative f'(x) = 4x.
Using the point-slope form of a linear equation, y - y₁ = m(x - x₁), where m represents the slope and (x₁, y₁) represents a point on the line, we can substitute the values:
y - 3 = 4( x - 1)
Expanding and simplifying the equation gives y - 3 = 4x - 4.
Hence, the equation of the tangent line to the curve at the point (1,3) is y = 4x - 1.
c. To sketch the graph of f(x) = 2x^2 + 1, you can follow these steps:
1. Plot a few key points on the graph. Choose different x-values and calculate the corresponding y-values using the given equation. For example, for x = -2, -1, 0, 1, and 2, calculate the corresponding y-values.
2. Connect the plotted points smoothly using a smooth curve. The curve should follow the general shape of a parabola, as the equation is quadratic.
3. Consider the opening direction of the parabola. In this case, because the coefficient of x^2 (2) is positive, the graph opens upward. This means that both ends of the parabola should point upwards.
4. Label the axes and any significant points on the graph, such as the vertex if there is one.
By following these steps, you should be able to sketch the graph of f(x) = 2x^2 + 1 accurately.
I hope this explanation helps you understand the problem better and allows you to complete your homework! Let me know if you have any further questions.