Calculus problem #26933

Calculus Level 1

In the above diagram, the red curve is the graph of a function $f,$ and the blue curve is the graph of its first derivative $f'.$ What is the relationship between $f'(-1)$ and $f''(1) ?$

f'(-1) < f''(1)

f'(-1) > f''(1)

f'(-1) = f''(1)

Relationship cannot be determined

4 solutions

Andrew Ellinor
Oct 29, 2015

Observe that $f'(-1)$ is a positive value since the graph of $f$ is increasing at $x=-1.$ Also observe that $f''(1)$ is a negative value since the graph of $f$ is concave down at $x=1.$ Thus, the value of $f'(-1)$ is larger than the value of $f''(1).$

But f"(x) isn't on the graph... shouldn't it be relationship cannot be determined because there is no apparent f"(x)?

Cindy Wang - 5 years, 2 months ago

$f''(x)$ is the derivative of $f'(x),$ so you can determine $f''(x)$ from $f'(x).$ In fact, you could solve this whole problem with just the graph of $f(x).$

Eli Ross Staff - 4 years, 8 months ago

How am I supposed to know if f '(-1) means the slope of f at -1 or the height of f '(-1)?

Alex Li - 5 years, 5 months ago

They mean the same thing

Andriy Mulyar - 5 years, 4 months ago

f'(x) at x=-1 and x=+1 are decreasing, consequently, f''(x) must be just negative at x=-1 and x=+1 and it's allowed to say f'(-1)>f''(1)

Gegham Asryan - 4 years, 11 months ago

Gurkirat Singh Chauhan
May 26, 2017

Graph of f' somewhat looks like cos x whose derivative is -sin x . And we know that cos( -1 ) > -sin ( 1 )

that is no solution just a lucky guess.. :-D

Jozko Mrkvicka - 3 years ago

7940 602300
Mar 1, 2018

The slope of the tangent line of the function f(x) at -1 is positive, and the slope of the function f'(x) at 1 is negative. So f'(-1)>f''(1).

Josiah Lee
Sep 25, 2019

f'(-1) is the value of y on the graph of f'(x) when x=-1 -> since value>0, f'(-1) is positive

f"(1) is the gradient of the graph of f'(x) when x=1 -> since the graph is decreasing at x=1, gradient is negative and f"(1) is negative.

Thus, f'(-1)>0>f"(1)

Calculus problem #26933

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