Suppose the function f(x) - f(2x) has the der | vedclass

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

Question

(A) Let $g(x) = f(x) - f(2x)$. Then $g'(x) = f'(x) - 2f'(2x)$.Given $g'(1) = 5$,we have $f'(1) - 2f'(2) = 5$  --- $(1)$.Given $g'(2) = 7$,we have $f'(

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(A) Let $g(x) = f(x) - f(2x)$. Then $g'(x) = f'(x) - 2f'(2x)$.Given $g'(1) = 5$,we have $f'(1) - 2f'(2) = 5$  --- $(1)$.Given $g'(2) = 7$,we have $f'(2) - 2f'(4) = 7$  --- $(2)$.We want to find the derivative of $h(x) = f(x) - f(4x)$ at $x = 1$.$h'(x) = f'(x) - 4f'(4x)$.At $x = 1$,$h'(1) = f'(1) - 4f'(4)$.From $(2)$,$f'(2) = 7 + 2f'(4)$.Substitute $f'(2)$ into $(1)$:$f'(1) - 2(7 + 2f'(4)) = 5$$f'(1) - 14 - 4f'(4) = 5$$f'(1) - 4f'(4) = 19$.Thus,$h'(1) = 19$.

Suppose the function $f(x) - f(2x)$ has the derivative $5$ at $x = 1$ and derivative $7$ at $x = 2$. The derivative of the function $f(x) - f(4x)$ at $x = 1$ has the value equal to:

Similar Questions

If $y=a \sin x+(5+2 x) \cos x$,then $y^{\prime \prime}+y=$

If $x^3 + y^3 = 6$ and $\frac{d^2y}{dx^2} \cdot \frac{d^2x}{dy^2} = \frac{m}{(xy)^n}$, where $m, n \in R$, then find the value of $\frac{m}{n}$.

Let $f(x)=ax^3+bx^2+cx+41$ be such that $f(1)=40, f'(1)=2$ and $f''(1)=4$. Then $a^2+b^2+c^2$ is equal to :

$A$ non-zero polynomial with real coefficients has the property that $f''(x) f'(x) = f(x)$. Then the value of $f'''(x)$ is

$f(x)$ is a twice differentiable function such that $f^{\prime \prime}(x)=-f(x)$ and $f^{\prime}(x)=g(x)$. If $h(x)=(f(x))^2+(g(x))^2$ and $h(1)=2$, then $h(2)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module