Hey everyone!! please help..
Find a function f such that the graph of f has a horizontal tangent at (1,0) and
f"(x)=10x??
Thanks
if f '' (x) =10x
then f ' (x) = 5x^2 + c
but at (1,0), f ' (x) = 0
0 = 5(1) + c
c = -5
so f ' (x) = 5x^2 - 5
f(x) = (5/3)x^3 - 5x + k
but (1,0) lies on it
0 = (5/3)(1) - 5 + k
k = 5 - 5/3 = 10/3
f(x) = (5/3)x^3 - 5x + 10/3
Why did the function go to therapy?
Because it couldn't seem to find its tangent! But no worries, I'm here to help you out. Let's find a function that satisfies your requirements.
Given that f"(x) = 10x, we can integrate this equation twice to find f(x). First, we integrate f"(x) to get f'(x):
f'(x) = 10x²/2 + C1
And now, let's integrate f'(x) to find f(x):
f(x) = (10x³)/6 + C1x + C2
To satisfy the condition of having a horizontal tangent at (1,0), we need the derivative, f'(x), to be equal to 0 at x = 1:
0 = 10(1)²/2 + C1
0 = 5 + C1
C1 = -5
Now our equation for f(x) becomes:
f(x) = (10x³)/6 - 5x + C2
So, the function that satisfies the given conditions is f(x) = (10x³)/6 - 5x + C2.
To find a function that satisfies the given conditions, we need to integrate the second derivative, f"(x), twice.
Given that f"(x) = 10x, we can integrate it once to find the first derivative, f'(x):
∫ f"(x) dx = ∫ 10x dx
Integrating 10x with respect to x gives us:
f'(x) = 5x^2 + C₁
where C₁ is the constant of integration.
Since we know that the graph of f has a horizontal tangent at (1,0), the slope of the tangent is 0. Therefore, f'(1) = 0:
f'(1) = 5(1)^2 + C₁ = 0
5 + C₁ = 0
C₁ = -5
Now we can find the original function f(x) by integrating f'(x):
∫ f'(x) dx = ∫ (5x^2 - 5) dx
Integrating (5x^2 - 5) with respect to x gives us:
f(x) = (5/3)x^3 - 5x + C₂
where C₂ is the constant of integration.
To find the constant C₂, we can use the given point (1,0). Substituting x = 1 and y = 0 into the equation of f(x):
f(1) = (5/3)(1)^3 - 5(1) + C₂ = 0
(5/3) - 5 + C₂ = 0
C₂ = 5 - (5/3) = 10/3
Therefore, the function f(x) that satisfies the given conditions is:
f(x) = (5/3)x^3 - 5x + (10/3).
To find a function f that satisfies the given conditions, we need to integrate the second derivative, f"(x) = 10x, twice to find f(x). The first integration will give us the first derivative, and the second integration will give us the function f(x).
Step 1: Integration to find the first derivative.
To find the first derivative, we integrate f"(x) = 10x with respect to x. Recall that when integrating a term with x raised to the power of n, you add 1 to the power and divide by the new power. In this case, we have x raised to the power of 1 (x^1) and the final power will be 2.
Integrating f"(x) = 10x:
∫ 10x dx = 10 * ∫ x dx
= 10 * (x^2 / 2) + C1
Where C1 is the constant of integration.
Step 2: Integration to find the function f(x).
Now that we have the first derivative, we need to find the function f(x). We integrate the first derivative with respect to x. In this case, we have (10 * (x^2 / 2)) as the integrand.
Integrating 10 * (x^2 / 2):
∫ 10 * (x^2 / 2) dx = (10/2) * ∫ x^2 dx
= 5 * (x^3 / 3) + C2
Where C2 is the constant of integration.
Step 3: Plugging in the given condition to find the constants C1 and C2.
To ensure that the graph of f has a horizontal tangent at (1, 0), we can substitute x = 1 and f(x) = 0 into the function f(x) we obtained.
Setting x = 1 and f(x) = 0 in the equation 5 * (x^3 / 3) + C2:
5 * (1^3 / 3) + C2 = 0
5/3 + C2 = 0
C2 = -5/3
We have found the constant C2.
Step 4: Finalizing the function f(x).
Using the constant C1 and C2 we found, we can finalize the function f(x). Plugging in the constants into our expression for f(x):
f(x) = 10 * (x^2 / 2) + C1
= 5 * x^2 + C1
Therefore, the function f(x) that satisfies the given conditions is:
f(x) = 5 * x^2 + C1
Note: The constant C1 can be any constant, as it does not affect the horizontal tangent.