How many critical points does the function f(x) = e^x - x^3 have?
wherever e^x - 3x^2 = 0
That is, where e^x = 3x^2
You might not know just where they are, but you do know that 3x^2 is a parabola with vertex at (0,0).
Since e^x is kind of flat near there, you know that for x<0 there will be an intersection. Now the question is, will there also be one where x>0?
at x=0, e^x = 1 > 3x^2 = 0
at x=1, e^x = e < 3x^2 = 3
So, somewhere in the interval (0,1) there will be another intersection.
at x=4, e^4 ? 54 > 3x^2 = 27
So, in the interval (1,4) there will be another intersection. After that e^x remains greater than 3x^2.
So, e^x-x^3 has three critical points, as seen at
http://www.wolframalpha.com/input/?i=e%5Ex+-+x%5E3+for+-1%3Cx%3C4
To find the critical points of a function, we need to find the values of x where the derivative of the function equals zero or is undefined.
Let's start by finding the derivative of f(x).
f(x) = e^x - x^3
f'(x) = d/dx(e^x) - d/dx(x^3)
The derivative of e^x is simply e^x.
f'(x) = e^x - 3x^2
Now we need to set the derivative equal to zero and solve for x.
e^x - 3x^2 = 0
We can't solve this equation algebraically, but we can find the critical points by using numerical methods or a graphing calculator.
Using a graphing calculator or a software, we find two critical points for this function.
Critical point 1: x ≈ -1.240
Critical point 2: x ≈ 2.219
Therefore, the function f(x) = e^x - x^3 has two critical points.
To determine the number of critical points of the function f(x) = e^x - x^3, we need to find the values of x where the derivative of f(x) is equal to zero or undefined. Let's begin by finding the derivative of f(x).
Step 1: Find the derivative of f(x)
f'(x) = (d/dx)[e^x - x^3]
Using the rules of differentiation, we can find the derivative of each term separately.
Derivative of e^x = e^x (since the derivative of e^x is equal to the original function)
Derivative of x^3 = 3x^2 (using the power rule of differentiation)
Therefore, f'(x) = e^x - 3x^2
Step 2: Set the derivative equal to zero and solve for x
To find the critical points, we set f'(x) = 0 and solve for x.
0 = e^x - 3x^2
Step 3: Solve the equation for x
To solve this equation, we need to use numerical or graphical methods since there is no algebraic solution. One possible method is to plot the graph of y = e^x - 3x^2 and observe where it crosses the x-axis. Another option is to use computational software or calculators that provide numerical solutions.
Using a graphing calculator or software, we find that the equation has two real solutions:
x ≈ -1.513 and x ≈ 0.293
These are the x-values where the derivative f'(x) is equal to zero.
Therefore, the function f(x) = e^x - x^3 has two critical points.