Let $f: \mathbb{R} \to \mathbb{R}$ be a polynomial function of degree four having extreme values at $x = 4$ and $x = 5$. If $\displaystyle \lim_{x \to 0} \frac{f(x)}{x^2} = 5$, then $f(2)$ is equal to:

Step 1: Define the polynomial and its derivative

We are given that $f(x)$ is a polynomial of degree four. We can write its general form as $f(x) = ax^4 + bx^3 + cx^2 + dx + e$. The extreme values occur where the first derivative $f'(x)$ is zero. Let's find the derivative of $f(x)$.

Step 2: Use the limit condition to find coefficients

We are given the limit $\lim_{x \to 0} \frac{f(x)}{x^2} = 5$. For this limit to be finite and non-zero, the terms $dx$ and $e$ in the numerator must be zero, otherwise the limit would be infinite or undefined. After setting $d=0$ and $e=0$, we can divide by $x^2$ and take the limit, which gives us $c=5$.

Step 3: Formulate the derivative based on extreme values

Since $f(x)$ has extreme values at $x=4$ and $x=5$, these are roots of $f'(x)=0$. Also, from the previous step, we know $d=0$, so $f'(0)=0$. This means $x=0$ is also a root of $f'(x)=0$. Therefore, $f'(x)$ must have factors $(x-4)$, $(x-5)$, and $x$. We can write $f'(x) = kx(x-4)(x-5)$ for some constant $k$. Expanding this, we get $f'(x) = k(x^3 - 9x^2 + 20x)$.

Step 4: Equate coefficients to find 'a' and 'b'

By comparing the coefficients of the two expressions for $f'(x)$, $4ax^3 + 3bx^2 + 10x$ and $k(x^3 - 9x^2 + 20x)$, we can solve for $k$, $a$, and $b$. From the coefficient of $x$, we find $k = 1/2$. Then, substituting $k$ back, we find $a = 1/8$ and $b = -3/2$.

Step 5: Write the complete polynomial function

Now that we have found the values of $a$, $b$, $c$, $d$, and $e$, we can write the complete polynomial function $f(x)$. Substituting $a=1/8$, $b=-3/2$, $c=5$, $d=0$, and $e=0$ into the general form, we get $f(x) = \frac{1}{8}x^4 - \frac{3}{2}x^3 + 5x^2$.

Step 6: Calculate f(2)

Finally, we need to find the value of $f(2)$. Substitute $x=2$ into the polynomial function we just found and simplify the expression.

More from this chapter

• Solve: Prove using Lagrange's Mean Value theorem. pi/3 - 1/(5sqrt(3)) > 3/5 > pi/3 - 1/8
• An empty inverted right circular conical vessel of base radius 12 cm and vertical height 18 cm is being filled with water at a constant rate of 12π cm³/s.At the
• Find the local maximum and minimum values of f(x) = x * sqrt(4 - x²).
• Find the maximum value of f(x) = 2x³ - 9x² + 12x + 5 on the interval [0, 3].
• Find the equation of the tangent and normal to the curve y = x³ - 3x + 2 at x = 1.
• Find the local maxima and local minima of the function f(x) = sin(x) + cos(x) on the interval [0, 2*pi].
• The rate of change of the area of a circle with respect to its radius r (in cm²/s), when r = 6 cm is:
• (a) Find the intervals in R for which the function f(x) = x^4 - 2x² is increasing or decreasing. OR (b) Find: ∫ (2x + 1)/(√(18 - 4x - x²)) dx.

All Application of Derivatives solutions →