If the graph of the function `y=f(x)` has a unique tangent at the point `(a ,0)` through which the graph passes, then evaluate `("lim")_(xveca)((log)_e{1+6f(x)})/(3f(x))`

If the graph of the function y=f(x) has a unique tangent at the point (a,0) through which the graph passes,then evaluate lim_(x rarr a)(log_(e){1+6f(x)})/(3f(x))

The graph of the function y = f(x) has a unique tangent at the point (e^a,0) through which the graph passes then lim__(x->e^a) (log_e {1+7f (x)} - sin f(x))/(3f(x))

The graph of the function y=f(x) has a unique tangent at the point (a,0) through which the graph passes then,lim_(x rarr a)((ln(1+6f(x)))/(3f(x))

If the graph of the function y=f(x) has a unique tangent at point (e^(a),0) (not vertical, non-horizontal) on the graph, then evaluate lim_(xtoe^(a))((1 +9f(x))-"tan"(f(x)))/(2f(x)) .

Draw the graph of the function: f(x)=log|x|

Statement-1: If the graph of the function y=f(x) has a unique tangent at the point (a, 0) through which lim_(x->a) log_e (1+6(f(x)))/(3f(x))=2 Statement-2: Since the graph passes through (a, 0). Therefore f(a) = 0. When f(a) = 0 given limit iszero by zero form so that it can be evaluate by using L-Hospital's rule.

Mark the graph of the function f (x) = 3e^(x + 5) - 7

If the graph of the function y = f(x) passes through the points (1, 2) and (3, 5), then int_(1)^(3)f'(x) dx=

If the graph of the function y=f(x) is as shown : the graph of y=(1)/(2)(|f(x)|-f(x)) is

Sketch the graph of the following function f(x)=|ln|x||