1. f(x)=x^2(e^x) find a point where the tangent is horizontal.
The answer is apparently (-2, 4/e^2). But I got 0 when I solved for the derivative. I don't get this.
2. What value of x does the function f(x)=(x^3+27)^(1/2) have vertical tangents?
My approach was to find undefined values, and so x basically couldn't be smaller than -27, yet the answer is -3. How?
f=x^2(e^x)
f' = x(x+2)e^x
Clearly, x=0 or x = -2
f(-2) = 4/e^2
f=(x^3+27)^(1/2)
f'= 3x^2/(2√(x^3+27))
the denominator is zero when x = -3
To find a point where the tangent to the graph of a function is horizontal, you need to set the derivative equal to 0 and solve for x. Let's go through your first question step by step.
1. f(x) = x^2(e^x)
To find the derivative of this function, you can use the product rule and the chain rule. The derivative of f(x) is given by:
f'(x) = 2x(e^x) + x^2(e^x)
2. Set the derivative equal to 0:
2x(e^x) + x^2(e^x) = 0
Now, we can factor out the common term (e^x) to simplify the equation:
e^x(2x + x^2) = 0
3. Now we have two possible solutions:
- Either e^x = 0 (which has no solution since e^x is always positive), or
- 2x + x^2 = 0
4. Solve the quadratic equation:
2x + x^2 = 0
This equation can be factored as:
x(x + 2) = 0
So, the possible values of x are x = 0 and x = -2.
5. Now that we have the potential x-values, substitute them back into the original function f(x) to find the corresponding y-values:
For x = 0:
f(0) = (0^2)(e^0) = 0(1) = 0
For x = -2:
f(-2) = (-2^2)(e^(-2)) = 4(e^(-2))
So, the point where the tangent is horizontal is (-2, 4/e^2). It seems there was an error in the answer you provided, as the correct y-coordinate should involve e^2, not e^(-2).
Now let's move on to your second question:
1. f(x) = (x^3 + 27)^(1/2)
To find the values of x where the function has vertical tangents, you need to find the x-values where the derivative is undefined.
2. Find the derivative of the function using the chain rule:
f'(x) = (1/2)(x^3 + 27)^(-1/2)(3x^2)
3. To find where the derivative is undefined, we set the denominator of the derivative equal to 0:
(x^3 + 27)^(-1/2) = 0
Recall that a fraction is undefined if the denominator is equal to 0.
4. Take note that (x^3 + 27)^(-1/2) will never be equal to 0 since it involves squares, and the square of any real number is always non-negative.
5. Therefore, there are no values of x that make the function have vertical tangents. The answer you provided, -3, seems to be incorrect.
I hope this explanation clarifies the process for finding horizontal and vertical tangents. If you have any further questions, feel free to ask!