Find values of a, b , c and d so that f(x) = ax^3 + bx^2 + cx + d has a relative minimum at (0,0) and a relati?
find f'(x)=0 at x=0,1 f'(x)=3ax^2 + 2bx + c f'(0)=0=c f'(1)=0=3a + 2b so: 3a=-2b a=(-2/3)b now: f(0)=0=d f(1)=1=(-2/3)b + b (1/3)b=1 b=3 so a=-2 so: f(x)=-2x^3+3x^2 graph to verify.
What is the minimum value of f(x)=e^x-2x?
f(x) = e^x - 2x f '(x) = e^x - 2 and f "(x) = e^x f '(x) = 0 when e^x - 2 = 0 e^x = 2 x = ln 2 When x = ln 2, f "(x) > 0 so local min occurs when x = ln 2 So y(min) = e^(ln 2) - 2 ln 2 = e^(ln 2) - ln 4 (which is approx 0.6137 to 4 dec pl.)
Find the values of a and b such that f(x) is differentiable at x = 2?
Hello - I haven't been able to figure out this question, any help is greatly appreciated! Let f(x) = bx^2 + 6x if x<=2 ax^3 if x>2 find the values of a and b such that f(x) is differentiable at x = 2. So far, I took the derivatives of each of the piecewise functions and plugged in 2. This left me with 4b+6 = 12a. How do I solve for each variable? Thanks!
How do I find maximum and minimum value of a function?
There are Various methods in order to find maximum or minimum value of a function. One of the conventional methods is:Find the derivative of the function and equate it to zero.Find the roots of the differentiated equation.Do double differentiation of original function and substitute the values of roots in the 2nd differentiated expression.If the value comes out to be negative, At the particular value of the root Maximum occurs. Then substitute the value in original expression to get Maximum of the function.If the value of double derivative after substituting the root is positive, Minimum occurs. Then substitute the value in original equation to get Minimum value of the function.If the Second derivative is Zero: Then go for higher derivatives of the function & substitute the value of the root in the nth order derivative expression. If it's positive it would give the Maximum of the function at the particular root.Hope the answer Helps.
Find the values of x for which f(x)=g(x)? (Professor never explained)?
Set f(x) = g(x) : x^2 + 2x + 1 = 3x + 3. Solve for all values of x.
How to find global extreme values of the this function ?
(i) Critical points inside the region. First, we find the first partial derivatives: f_x = 12x^2 + 8xy = 4x(3x + 2y) f_y = 4x^2 + 6y = 2(2x^2 + 3y). Set these equal to 0: 4x(3x + 2y) = 0 ==> x = 0 or y = -3x/2. 2(2x^2 + 3y) = 0 ==> 2x^2 + 3y = 0. If x = 0, then 0 + 3y = 0 ==> y = 0. If y = -3x/2, then 2x^2 + 3 * (-3x/2) = 0. ==> x(4x - 9) = 0. ==> x = 0 or x = 9/4 (ignore x = 9/4, since this is not in the region.) So, the only critical point is (0, 0). (Being on the boundary of D, we can ignore this as well; it will arise again below.) ------ (ii) Extreme values on the boundary of the region, which is bounded by x = 0, y = 0, and x + y = 1. Examine this one edge at a time. (a) x = 0 with y in [0, 1]: f(0, y) = 3y^2, which has extrema 0 (at y = 0) and 3 (at y = 1). (b) y = 0 with x in [0, 1]: f(x, 0) = 4x^3, which has extrema 0 (at x = 0) and 4 (at x = 1). (c) x + y = 1 <==> y = 1 - x with x in [0, 1]: g(x) := f(x, 1-x) ........= 4x^3 + 4x^2(1 - x) + 3(1 - x)^2 ........= 7x^2 - 6x + 3. g'(x) = 14x - 6. Setting this equal to 0 yields x = 3/7, which is in [0, 1]. Substitute this (and the endpoints) into g: g(0) = 3 g(1) = 4 g(3/7) = 12/7. -------------- Overall, the extreme values of f on the region are 0 and 4. I hope this helps!
What is the minimum and maximum value of q= cos(2x) +2cos(x)?
Finding out the maxima in this case is pretty intuitive. We know that cosine of [math]x[/math] attains the maximum value of [math]1[/math] at [math]x = 0[/math]. If [math]x = 0[/math], [math]2x[/math] also [math]= 0[/math], so, we can directly conclude the maximum value of the given function as [math]1 + 2 \times 1 = 3[/math]For the minimum value, I believe the derivatives approach will have to employed[math]q = cos(2x) +2cos(x)[/math][math]\dfrac{dq}{dx} = -2sin(2x) -2sin(x) = 0[/math]or [math]sin(2x) + sin(x) = 0[/math]or [math]sin(x)[2 cos(x) + 1] = 0[/math]so, [math]sin(x) = 0[/math] or [math]cos(x) = -1/2[/math]The second solution can be used to arrive at the value [math]-1.5[/math]. You should try it out yourself, and let me know if you face any issue.
For the 2 functions, f(x) and g(x), tables of values are shown below. What is the value of g(f(3))?
Hello First, you find the value for f(3), which is listed to be 2. Since the value of f(3) is 2, it's equivalent to you finding the value of g(2), which is listed to be -3. Therefore the answer is B. hope this helps! :)