Verify Rolle’s theorem for the function f (x) = x2 + 2x – 8, x �

1.	Verify Rolle’s theorem for the function f (x) = x2 + 2x – 8, x ∈ [– 4, 2].
Answer» Given function is f(x) = x² +2 x − 8, x ∈ [ −4, 2] Since, a polynomial function is continuous and derivable on R, therefore (i) f(x) is continuous on [− 4,2]. (ii) f(x) is derivable on (− 4,2). Also, f(− 4) = (− 4)² + 2 (− 4) − 8 = 0 (since f (x) = x² + 2x − 8) and f(2) = 2² + 2 x 2 – 8 = 0 ⇒ f(− 4) = f(2) This means that all the conditions of Rolle's theorem are satisfied by f(x) in [− 4,2]. Therefore, it exists at least one real c ∈ [ −4, 2] such that f′(c) = 0. Now, f (x) = x² + 2x − 8 ⇒ f'(x) = d/dx(x² + 2x - 8) = 2x + 2 Putting f′(c) = 0 ⇒ 2c + 2 = 0 ⇒ c = − 1. Thus, f′( −1) =0 and − 1 ∈ (− 4,2). Hence, Rolle's theorem is verified with c = − 1.

Verify Rolle’s theorem for the function f (x) = x2 + 2x – 8, x ∈ [– 4, 2].

Answer»

Given function is f(x) = x² +2 x − 8, x ∈ [ −4, 2]

Since, a polynomial function is continuous and derivable on R, therefore

(i) f(x) is continuous on [− 4,2].

(ii) f(x) is derivable on (− 4,2).

Also, f(− 4) = (− 4)² + 2 (− 4) − 8 = 0

(since f (x) = x² + 2x − 8)

and f(2) = 2² + 2 x 2 – 8 = 0

⇒ f(− 4) = f(2)

This means that all the conditions of Rolle's theorem are satisfied by f(x) in [− 4,2].

Therefore, it exists at least one real c ∈ [ −4, 2] such that f′(c) = 0.

Now, f (x) = x² + 2x − 8 ⇒ f'(x) = d/dx(x² + 2x - 8) = 2x + 2

Putting f′(c) = 0 ⇒ 2c + 2 = 0 ⇒ c = − 1.

Thus, f′( −1) =0 and − 1 ∈ (− 4,2).

Hence, Rolle's theorem is verified with c = − 1.

Discussion

No Comment Found

Related InterviewSolutions

If `f(x)=2x-|x|`, then at `x=0`A. f is continuousB. f is discontinuousC. `underset(x rarr 0^(-))lim f(x)=3`D. `underset(x rarr 0^(+))lim f(x)=1`
If `f(x)={:{((x)/(|x|)", for " x!=0),(c", for " x=0):}`, then f isA. continuous at `x=0`B. discontinuous at `x=0`C. continuous if `f(0)=1`D. continuous if `f(0)=-1`
Identify discontinuities if any for the following functions as either a jump or a removable discontinuity on their respective domains.f(x) = x2 + 5x + 1, for 0 ≤ x ≤ 3 = x2 + x + 5, for 3 < x ≤ 6
If `f(x)=a^(x), a gt 0`, then f isA. continuous for all `x in R^(+)`B. continuous for all `x in R^(-)`C. continuous for all ` x in R`D. discontinuous for all ` x in R`
The rational function `f(x)=(g(x))/(h(x)), h(x) != 0` isA. continuousB. discontinuous for integer values onlyC. continuous for integer values onlyD. continuous for imaginary values only
Discuss the continuity of the function `log_c x,` where `c> 0, x > 0.`A. continuous in `(-oo, oo)`B. continuous in `(0, oo)`C. discontinuous in `(-oo, oo)`D. discontinuous in `(0, oo)`
Examine the continuity of the given function at given points `f(x)=(x+3x^2+5x^3+.....+(2n-1)x^n-n^2)/(x-1)`, for `x != 1` at `x=1 and =(n(n^2-1))/3`,for `x=1`A. `(n(n+1)(2n-1))/(6)`B. `(n(n+1)(2n-1))/(3)`C. `(n(n+1)(4n-1))/(6)`D. `(n(n+1)(4n-1))/(3)`
If \(f(x) =\begin{cases}x\,sin \frac{1}{x}, & \quad x\neq 0\\k, & \quad x=0\end{cases}\) is continuous at x = 0, then the value of k =f(x) = {x sin(1/x), x ≠ 0, k, x = 0 is continuous at x = 0, then the value of k(A) 1(B) –1(C) 0(D) 2
Prove that \(f(x) = \begin{cases} 2-x, & \quad \text{when x <2;} \text{}\\ 2+x, & \quad \text{whenx≥2} \end{cases}\) is discontinuous at x=2
Prove that \(f(x) = \begin{cases} 3-x, & \quad \text{when x ≤0;} \text{}\\ x^2, & \quad \text{whex>0} \end{cases}\) is discontinuous at x = 0