If f(x)=x^2+ax+b has a minima at x=3 whose va | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the function $f(x) = x^2 + ax + b$. To find the critical points,we calculate the first derivative: $f'(x) = 2x + a$. Setting $f'(x) = 0$,we

Vedclass · Accepted Answer

-$6$,$14$

Vedclass · Answer

(A) Given the function $f(x) = x^2 + ax + b$. To find the critical points,we calculate the first derivative: $f'(x) = 2x + a$. Setting $f'(x) = 0$,we get $x = -a/2$. Since the second derivative $f''(x) = 2 > 0$,the function has a local minimum at $x = -a/2$. According to the problem,the minimum occurs at $x = 3$. Therefore,$-a/2 = 3$,which implies $a = -6$. Given that the minimum value of the function is $5$ at $x = 3$,we substitute these values into the original function: $f(3) = (3)^2 + a(3) + b = 5$. Substituting $a = -6$: $9 + (-6)(3) + b = 5$. $9 - 18 + b = 5$. $-9 + b = 5$. $b = 14$. Thus,the values are $a = -6$ and $b = 14$.

If $f(x)=x^2+ax+b$ has a minima at $x=3$ whose value is $5$,then the values of $a$ and $b$ are respectively.

Similar Questions

The function $f(x)=2 x^3-9 a x^2+12 a^2 x+1$ $(a>0)$ attains its maximum and minimum at $p$ and $q$ respectively and $p^2=q$. Then,$a=$

The global maximum value of $f(x) = \log_{10}(4x^3 - 12x^2 + 11x - 3)$,$x \in [2, 3]$,is

If $f(x) = x^5 - 5x^4 + 5x^3 - 10$ has local maximum and minimum at $x = p$ and $x = q$,respectively,then $(p, q)$ equals

If $y=a \log x+b x^2+x$ has its extreme value at $x=-1$ and $x=2$,then the value of $a+b$ is

If the absolute maximum value of the function $f(x) = (x^2 - 2x + 7) e^{(4x^3 - 12x^2 - 180x + 31)}$ in the interval $[-3, 0]$ is $f(\alpha)$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module